di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Geometry


Question: Let $A B C D$ be a quadrilateral inscribed in a circle $\omega$. The lines $A B$ and $C D$ meet at $P$, the lines $A D$ and $B C$ meet at $Q$, and the diagonals $A C$ and $B D$ meet at $R$. Let $M$ be the midpoint of the segment $P Q$, and let $K$ be the common point of the segment $M R$ and the circle $\omega$. Prove that the circumcircle of the triangle $K P Q$ and $\omega$ are tangent to one another.

Solution:
Let $O$ be the centre of $\omega$. Notice that the points $P, Q$, and $R$ are the poles (with respect to $\omega$ ) of the lines $Q R, R P$, and $P Q$, respectively. Hence we have $O P \perp Q R, O Q \perp R P$, and $O R \perp P Q$, thus $R$ is the orthocentre of the triangle $O P Q$. Now, if $M R \perp P Q$, then the points $P$ and $Q$ are the reflections of one another in the line $M R=M O$, and the triangle $P Q K$ is symmetrical with respect to this line. In this case the statement of the problem is trivial.



Otherwise, let $V$ be the foot of the perpendicular from $O$ to $M R$, and let $U$ be the common point of the lines $O V$ and $P Q$. Since $U$ lies on the polar line of $R$ and $O U \perp M R$, we obtain that $U$ is the pole of $M R$. Therefore, the line $U K$ is tangent to $\omega$. Hence it is enough to prove that $U K^{2}=U P \cdot U Q$, since this relation implies that $U K$ is also tangent to the circle $K P Q$.



From the rectangular triangle $O K U$, we get $U K^{2}=U V \cdot U O$. Let $\Omega$ be the circumcircle of triangle $O P Q$, and let $R^{\prime}$ be the reflection of its orthocentre $R$ in the midpoint $M$ of the side $P Q$. It is well known that $R^{\prime}$ is the point of $\Omega$ opposite to $O$, hence $O R^{\prime}$ is the diameter of $\Omega$. Finally, since $\angle O V R^{\prime}=90^{\circ}$, the point $V$ also lies on $\Omega$, hence $U P \cdot U Q=U V \cdot U O=U K^{2}$, as required.



<img_3711>
Subfield: Algebra


Question: Does there exist an infinite sequence of positive integers $a_{1}, a_{2}, a_{3}, \ldots$ such that $a_{m}$ and $a_{n}$ are coprime if and only if $|m-n|=1$ ?

Solution:
The answer is in the affirmative.



The idea is to consider a sequence of pairwise distinct primes $p_{1}, p_{2}, p_{3}, \ldots$, cover the positive integers by a sequence of finite non-empty sets $I_{n}$ such that $I_{m}$ and $I_{n}$ are disjoint if and only if $m$ and $n$ are one unit apart, and set $a_{n}=\prod_{i \in I_{n}} p_{i}, n=1,2,3, \ldots$



One possible way of finding such sets is the following. For all positive integers $n$, let



$$

\begin{aligned}

& 2 n \in I_{k} \quad \text { for all } k=n, n+3, n+5, n+7, \ldots ; \quad \text { and } \\

& 2 n-1 \in I_{k} \quad \text { for all } k=n, n+2, n+4, n+6, \ldots

\end{aligned}

$$



Clearly, each $I_{k}$ is finite, since it contains none of the numbers greater than $2 k$. Next, the number $p_{2 n}$ ensures that $I_{n}$ has a common element with each $I_{n+2 i}$, while the number $p_{2 n-1}$ ensures that $I_{n}$ has a common element with each $I_{n+2 i+1}$ for $i=1,2, \ldots$. Finally, none of the indices appears in two consecutive sets.
Subfield: Geometry


Question: Fix a circle $\Gamma$, a line $\ell$ tangent to $\Gamma$, and another circle $\Omega$ disjoint from $\ell$ such that $\Gamma$ and $\Omega$ lie on opposite sides of $\ell$. The tangents to $\Gamma$ from a variable point $X$ on $\Omega$ cross $\ell$ at $Y$ and $Z$. Prove that, as $X$ traces $\Omega$, the circle $X Y Z$ is tangent to two fixed circles.

Solution:
Assume $\Gamma$ of unit radius and invert with respect to $\Gamma$. No reference will be made to the original configuration, so images will be denoted by the same letters. Letting $\Gamma$ be centred at $G$, notice that inversion in $\Gamma$ maps tangents to $\Gamma$ to circles of unit diameter through $G$ (hence internally tangent to $\Gamma$ ). Under inversion, the statement reads as follows:

Fix a circle $\Gamma$ of unit radius centred at $G$, a circle $\ell$ of unit diameter through $G$, and a circle $\Omega$ inside $\ell$ disjoint from $\ell$. The circles $\eta$ and $\zeta$ of unit diameter, through $G$ and a variable point $X$ on $\Omega$, cross $\ell$ again at $Y$ and $Z$, respectively. Prove that, as $X$ traces $\Omega$, the circle $X Y Z$ is tangent to two fixed circles.

<img_3955>

Since $\eta$ and $\zeta$ are the reflections of the circumcircle $\ell$ of the triangle $G Y Z$ in its sidelines $G Y$ and $G Z$, respectively, they pass through the orthocentre of this triangle. And since $\eta$ and $\zeta$ cross again at $X$, the latter is the orthocentre of the triangle $G Y Z$. Hence the circle $\xi$ through $X, Y, Z$ is the reflection of $\ell$ in the line $Y Z$; in particular, $\xi$ is also of unit diameter.

Let $O$ and $L$ be the centres of $\Omega$ and $\ell$, respectively, and let $R$ be the (variable) centre of $\xi$. Let $G X \operatorname{cross} \xi$ again at $X^{\prime}$; then $G$ and $X^{\prime}$ are reflections of one another in the line $Y Z$, so $G L R X^{\prime}$ is an isosceles trapezoid. Then $L R \| G X$ and $\angle(L G, G X)=\angle\left(G X^{\prime}, X^{\prime} R\right)=\angle(R X, X G)$, i.e., $L G \| R X$; this means that $G L R X$ is a parallelogram, so $\overrightarrow{X R}=\overrightarrow{G L}$ is constant.

Finally, consider the fixed point $N$ defined by $\overrightarrow{O N}=\overrightarrow{G L}$. Then $X R N O$ is a parallelogram, so the distance $R N=O X$ is constant. Consequently, $\xi$ is tangent to the fixed circles centred at $N$ of radii $|1 / 2-O X|$ and $1 / 2+O X$.

One last check is needed to show that the inverse images of the two obtained circles are indeed circles and not lines. This might happen if one of them contained $G$; we show that this is impossible. Indeed, since $\Omega$ lies inside $\ell$, we have $O L<1 / 2-O X$, so

$$
N G=|\overrightarrow{G L}+\overrightarrow{L O}+\overrightarrow{O N}|=|2 \overrightarrow{G L}+\overrightarrow{L O}| \geq 2|\overrightarrow{G L}|-|\overrightarrow{L O}|>1-(1 / 2-O X)=1 / 2+O X
$$

this shows that $G$ is necessarily outside the obtained circles.
Subfield: Algebra


Question: A cubic sequence is a sequence of integers given by $a_{n}=n^{3}+b n^{2}+c n+d$, where $b, c$ and $d$ are integer constants and $n$ ranges over all integers, including negative integers.

Determine the possible values of $a_{2015} \cdot a_{2016}$ for a cubic sequence satisfying the condition in part (a).

Solution:
The only possible value of $a_{2015} \cdot a_{2016}$ is 0 . For simplicity, by performing a translation of the sequence (which may change the defining constants $b, c$ and $d$ ), we may instead concern ourselves with the values $a_{0}$ and $a_{1}$, rather than $a_{2015}$ and $a_{2016}$.



Suppose now that we have a cubic sequence $a_{n}$ with $a_{0}=p^{2}$ and $a_{1}=q^{2}$ square numbers. We will show that $p=0$ or $q=0$. Consider the line $y=(q-p) x+p$ passing through $(0, p)$ and $(1, q)$; the latter are two points the line under consideration and the cubic $y^{2}=x^{3}+b x^{2}+c x+d$ share. Hence the two must share a third point whose $x$-coordinate is the third root of the polynomial $t^{3}+\left(b-(q-p)^{2}\right) t^{2}+(c-2(q-p) p) t+\left(d-p^{2}\right)$ (it may well happen that this third point coincide with one of the other two points the line and the cubic share).



Notice that the sum of the three roots is $(q-p)^{2}-b$, so the third intersection has integral $x$-coordinate $X=(q-p)^{2}-b-1$. Its $y$-coordinate $Y=(q-p) X+p$ is also an integer, and hence $a_{X}=X^{3}+b X^{2}+c X+d=Y^{2}$ is a square. This contradicts our assumption on the sequence unless $X=0$ or $X=1$, i.e. unless $(q-p)^{2}=b+1$ or $(q-p)^{2}=b+2$.







Applying the same argument to the line through $(0,-p)$ and $(1, q)$, we find that $(q+p)^{2}=b+1$ or $b+2$ also. Since $(q-p)^{2}$ and $(q+p)^{2}$ have the same parity, they must be equal, and hence $p q=0$, as desired.



It remains to show that such sequences exist, say when $p=0$. Consider the sequence $a_{n}=$ $n^{3}+\left(q^{2}-2\right) n^{2}+n$, chosen to satisfy $a_{0}=0$ and $a_{1}=q^{2}$. We will show that when $q=1$, the only square terms of the sequence are $a_{0}=0$ and $a_{1}=1$. Indeed, suppose that $a_{n}=n\left(n^{2}-n+1\right)$ is square. Since the second factor is positive, and the two factors are coprime, both must be squares; in particular, $n \geq 0$. The case $n=0$ is clear, so let $n \geq 1$. Finally, if $n>1$, then $(n-1)^{2}<n^{2}-n+1<n^{2}$, so $n^{2}-n+1$ is not a square. Consequently, $n=0$ or $n=1$, and the conclusion follows.

Final answer: 0
Subfield: Combinatorics


Question: Amy and Bob play the game. At the beginning, Amy writes down a positive integer on the board. Then the players take moves in turn, Bob moves first. On any move of his, Bob replaces the number $n$ on the blackboard with a number of the form $n-a^{2}$, where $a$ is a positive integer. On any move of hers, Amy replaces the number $n$ on the blackboard with a number of the form $n^{k}$, where $k$ is a positive integer. Bob wins if the number on the board becomes zero. Can Amy prevent Bob's win?

Solution:
The answer is in the negative. For a positive integer $n$, we define its square-free part $S(n)$ to be the smallest positive integer $a$ such that $n / a$ is a square of an integer. In other words, $S(n)$ is the product of all primes having odd exponents in the prime expansion of $n$. We also agree that $S(0)=0$.



Now we show that (i) on any move of hers, Amy does not increase the square-free part of the positive integer on the board; and (ii) on any move of his, Bob always can replace a positive integer $n$ with a non-negative integer $k$ with $S(k)<S(n)$. Thus, if the game starts by a positive integer $N$, Bob can win in at most $S(N)$ moves.



Part (i) is trivial, as the definition of the square-part yields $S\left(n^{k}\right)=S(n)$ whenever $k$ is odd, and $S\left(n^{k}\right)=1 \leq S(n)$ whenever $k$ is even, for any positive integer $n$.



Part (ii) is also easy: if, before Bob's move, the board contains a number $n=S(n) \cdot b^{2}$, then Bob may replace it with $n^{\prime}=n-b^{2}=(S(n)-1) b^{2}$, whence $S\left(n^{\prime}\right) \leq S(n)-1$.
Subfield: Algebra


Question: Prove that there exist two functions



$$
f, g: \mathbb{R} \rightarrow \mathbb{R}
$$



such that $f \circ g$ is strictly decreasing, while $g \circ f$ is strictly increasing.

Solution:
Let



$$

\begin{aligned}

& \text { - } A=\bigcup_{k \in \mathbb{Z}}\left(\left[-2^{2 k+1},-2^{2 k}\right) \bigcup\left(2^{2 k}, 2^{2 k+1}\right]\right) ; \\

& \text { - } B=\bigcup_{k \in \mathbb{Z}}\left(\left[-2^{2 k},-2^{2 k-1}\right) \bigcup\left(2^{2 k-1}, 2^{2 k}\right]\right) .

\end{aligned}

$$



Thus $A=2 B, B=2 A, A=-A, B=-B, A \cap B=\varnothing$, and finally $A \cup B \cup\{0\}=\mathbb{R}$. Let us take



$$

f(x)=\left\{\begin{array}{lll}

x & \text { for } & x \in A \\

-x & \text { for } & x \in B \\

0 & \text { for } & x=0

\end{array}\right.

$$



Take $g(x)=2 f(x)$. Thus $f(g(x))=f(2 f(x))=-2 x$ and $g(f(x))=2 f(f(x))=2 x$.
Subfield: Geometry


Question: Let $A B C D$ be a cyclic quadrangle and let $P$ be a point on the side $A B$. The diagonal $A C$ crosses the segment $D P$ at $Q$. The parallel through $P$ to $C D$ crosses the extension of the side $B C$ beyond $B$ at $K$, and the parallel through $Q$ to $B D$ crosses the extension of the side $B C$ beyond $B$ at $L$. Prove that the circumcircles of the triangles $B K P$ and $C L Q$ are tangent.

Solution:
We show that the circles $B K P$ and $C L Q$ are tangent at the point $T$ where the line $D P$ crosses the circle $A B C D$ again.

Since $B C D T$ is cyclic, we have $\angle K B T=\angle C D T$. Since $K P \| C D$, we get $\angle C D T=\angle K P T$. Thus, $\angle K B T=\angle C D T=\angle K P T$, which shows that $T$ lies on the circle $B K P$. Similarly, the equalities $\angle L C T=\angle B D T=\angle L Q T$ show that $T$ also lies on the circle $C L Q$.

It remains to prove that these circles are indeed tangent at $T$. This follows from the fact that the chords $T P$ and $T Q$ in the circles $B K T P$ and $C L T Q$, respectively, both lie along the same line and subtend equal angles $\angle T B P=\angle T B A=\angle T C A=\angle T C Q$.

<img_3179>

### 分析
Subfield: Algebra


Question: Determine whether there exist non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying
$$
P(x)^{10}+P(x)^{9}=Q(x)^{21}+Q(x)^{20}
$$

Solution:
The answer is in the negative. Comparing the degrees of both sides in $(1)$ we get $\operatorname{deg} P=21 n$ and $\operatorname{deg} Q=10 n$ for some positive integer $n$. Take the derivative of $(1)$ to obtain

$$
P^{\prime} P^{8}(10 P+9)=Q^{\prime} Q^{19}(21 Q+20)
$$

Since $\operatorname{gcd}(10 P+9, P)=\operatorname{gcd}(10 P+9, P+1)=1$, it follows that $\operatorname{gcd}\left(10 P+9, P^{9}(P+1)\right)=1$, so $\operatorname{gcd}(10 P+9, Q)=1$, by $(1)$. Thus $(2)$ yields $10 P+9 \mid Q^{\prime}(21 Q+20)$, which is impossible since $0<\operatorname{deg}\left(Q^{\prime}(21 Q+20)\right)=20 n-1<21 n=\operatorname{deg}(10 P+9)$. A contradiction.
Letting $r$ and $s$ be integers such that $r \geq 2$ and $s \geq 2 r$, we show that if $P^{r}+P^{r-1}=$ $Q^{s}+Q^{s-1}$, then $Q$ is constant.

Let $m=\operatorname{deg} P$ and $n=\operatorname{deg} Q$. A degree inspection in the given relation shows that $m \geq 2 n$.

We will prove that $P(P+1)$ has at least $m+1$ distinct complex roots. Assuming this for the moment, notice that $Q$ takes on one of the values 0 or -1 at each of those roots. Since $m+1 \geq 2 n+1$, it follows that $Q$ takes on one of the values 0 and -1 at more than $n$ distinct points, so $Q$ must be constant.

Finally, we prove that $P(P+1)$ has at least $m+1$ distinct complex roots. This can be done either by referring to the Mason-Stothers theorem or directly, in terms of multiplicities of the roots in question.

Since $P$ and $P+1$ are relatively prime, the Mason-Stothers theorem implies that the number of distinct roots of $P(P+1)$ is greater than $m$, hence at least $m+1$.

For a direct proof, let $z_{1}, \ldots, z_{t}$ be the distinct complex roots of $P(P+1)$, and let $z_{k}$ have multiplicity $\alpha_{k}, k=1, \ldots, t$. Since $P$ and $P+1$ have no roots in common, and $P^{\prime}=(P+1)^{\prime}$, it follows that $P^{\prime}$ has a root of multiplicity $\alpha_{k}-1$ at $z_{k}$. Consequently, $m-1=\operatorname{deg} P^{\prime} \geq$ $\sum_{k=1}^{t}\left(\alpha_{k}-1\right)=\sum_{k=1}^{t} \alpha_{k}-t=2 m-t$; that is, $t \geq m+1$. This completes the prof.

### 分析
Subfield: Combinatorics


Question: Ann and Bob play a game on an infinite checkered plane making moves in turn; Ann makes the first move. A move consists in orienting any unit grid-segment that has not been oriented before. If at some stage some oriented segments form an oriented cycle, Bob wins. Does Bob have a strategy that guarantees him to win?

Solution:
The answer is in the negative: Ann has a strategy allowing her to prevent Bob's victory.

We say that two unit grid-segments form a low-left corner (or LL-corner) if they share an endpoint which is the lowest point of one and the leftmost point of the other. An up-right corner (or UR-corner) is defined similarly. The common endpoint of two unit grid-segments at a corner is the joint of that corner.

Fix a vertical line on the grid and call it the midline; the unit grid-segments along the midline are called middle segments. The unit grid-segments lying to the left/right of the midline are called left/right segments. Partition all left segments into LL-corners, and all right segments into UR-corners.

We now describe Ann's strategy. Her first move consists in orienting some middle segment arbitrarily. Assume that at some stage, Bob orients some segment $s$. If $s$ is a middle segment, Ann orients any free middle segment arbitrarily. Otherwise, $s$ forms a corner in the partition with some other segment $t$. Then Ann orients $t$ so that the joint of the corner is either the source of both arrows, or the target of both. Notice that after any move of Ann's, each corner in the partition is either completely oriented or completely not oriented. This means that Ann can always make a required move.

Assume that Bob wins at some stage, i.e., an oriented cycle $C$ occurs. Let $X$ be the lowest of the leftmost points of $C$, and let $Y$ be the topmost of the rightmost points of $C$. If $X$ lies (strictly) to the left of the midline, then $X$ is the joint of some corner whose segments are both oriented. But, according to Ann's strategy, they are oriented so that they cannot occur in a cycle - a contradiction. Otherwise, $Y$ lies to the right of the midline, and a similar argument applies. Thus, Bob will never win, as desired.


### 分析
Subfield: Algebra


Question: Given a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of real numbers. For each $i(1 \leq i \leq n)$ define

$$
d_{i}=\max \left\{a_{j}: 1 \leq j \leq i\right\}-\min \left\{a_{j}: i \leq j \leq n\right\}
$$

and let

$$
d=\max \left\{d_{i}: 1 \leq i \leq n\right\}
$$
Prove that for arbitrary real numbers $x_{1} \leq x_{2} \leq \ldots \leq x_{n}$,
$$
\max \left\{\left|x_{i}-a_{i}\right|: 1 \leq i \leq n\right\} \geq \frac{d}{2} \tag{1}
$$

Solution:
Let $1 \leq p \leq q \leq r \leq n$ be indices for which

$$
d=d_{q}, \quad a_{p}=\max \left\{a_{j}: 1 \leq j \leq q\right\}, \quad a_{r}=\min \left\{a_{j}: q \leq j \leq n\right\}
$$

and thus $d=a_{p}-a_{r}$. (These indices are not necessarily unique.)

<img_3891>

For arbitrary real numbers $x_{1} \leq x_{2} \leq \ldots \leq x_{n}$, consider just the two quantities $\left|x_{p}-a_{p}\right|$ and $\left|x_{r}-a_{r}\right|$. Since

$$
\left(a_{p}-x_{p}\right)+\left(x_{r}-a_{r}\right)=\left(a_{p}-a_{r}\right)+\left(x_{r}-x_{p}\right) \geq a_{p}-a_{r}=d
$$

we have either $a_{p}-x_{p} \geq \frac{d}{2}$ or $x_{r}-a_{r} \geq \frac{d}{2}$. Hence,

$$
\max \left\{\left|x_{i}-a_{i}\right|: 1 \leq i \leq n\right\} \geq \max \left\{\left|x_{p}-a_{p}\right|,\left|x_{r}-a_{r}\right|\right\} \geq \max \left\{a_{p}-x_{p}, x_{r}-a_{r}\right\} \geq \frac{d}{2}
$$
Subfield: Algebra


Question: Given a sequence $a_{1}, a_{2}, \ldots, a_{n}$ of real numbers. For each $i(1 \leq i \leq n)$ define

$$
d_{i}=\max \left\{a_{j}: 1 \leq j \leq i\right\}-\min \left\{a_{j}: i \leq j \leq n\right\}
$$

and let

$$
d=\max \left\{d_{i}: 1 \leq i \leq n\right\}
$$
Show that there exists a sequence $x_{1} \leq x_{2} \leq \ldots \leq x_{n}$ of real numbers such that we have equality in (1).

Solution:
Define the sequence $\left(x_{k}\right)$ as

$$
x_{1}=a_{1}-\frac{d}{2}, \quad x_{k}=\max \left\{x_{k-1}, a_{k}-\frac{d}{2}\right\} \quad \text { for } 2 \leq k \leq n
$$

We show that we have equality in (1) for this sequence.

By the definition, sequence $\left(x_{k}\right)$ is non-decreasing and $x_{k}-a_{k} \geq-\frac{d}{2}$ for all $1 \leq k \leq n$. Next we prove that

$$
x_{k}-a_{k} \leq \frac{d}{2} \quad \text { for all } 1 \leq k \leq n\tag{2}
$$

Consider an arbitrary index $1 \leq k \leq n$. Let $\ell \leq k$ be the smallest index such that $x_{k}=x_{\ell}$. We have either $\ell=1$, or $\ell \geq 2$ and $x_{\ell}>x_{\ell-1}$. In both cases,

$$
x_{k}=x_{\ell}=a_{\ell}-\frac{d}{2}\tag{3}
$$

Since

$$
a_{\ell}-a_{k} \leq \max \left\{a_{j}: 1 \leq j \leq k\right\}-\min \left\{a_{j}: k \leq j \leq n\right\}=d_{k} \leq d
$$

equality (3) implies

$$
x_{k}-a_{k}=a_{\ell}-a_{k}-\frac{d}{2} \leq d-\frac{d}{2}=\frac{d}{2}
$$

We obtained that $-\frac{d}{2} \leq x_{k}-a_{k} \leq \frac{d}{2}$ for all $1 \leq k \leq n$, so

$$
\max \left\{\left|x_{i}-a_{i}\right|: 1 \leq i \leq n\right\} \leq \frac{d}{2}
$$

We have equality because $\left|x_{1}-a_{1}\right|=\frac{d}{2}$.
Subfield: Algebra


Question: Consider those functions $f: \mathbb{N} \rightarrow \mathbb{N}$ which satisfy the condition

$$
f(m+n) \geq f(m)+f(f(n))-1\tag{1}
$$

for all $m, n \in \mathbb{N}$. Find all possible values of $f(2007)$.

( $\mathbb{N}$ denotes the set of all positive integers.)

Solution:
Suppose that a function $f: \mathbb{N} \rightarrow \mathbb{N}$ satisfies (1). For arbitrary positive integers $m>n$, by (1) we have

$$
f(m)=f(n+(m-n)) \geq f(n)+f(f(m-n))-1 \geq f(n),
$$

so $f$ is nondecreasing.

Function $f \equiv 1$ is an obvious solution. To find other solutions, assume that $f \not \equiv 1$ and take the smallest $a \in \mathbb{N}$ such that $f(a)>1$. Then $f(b) \geq f(a)>1$ for all integer $b \geq a$.

Suppose that $f(n)>n$ for some $n \in \mathbb{N}$. Then we have

$$
f(f(n))=f((f(n)-n)+n) \geq f(f(n)-n)+f(f(n))-1
$$

so $f(f(n)-n) \leq 1$ and hence $f(n)-n<a$. Then there exists a maximal value of the expression $f(n)-n$; denote this value by $c$, and let $f(k)-k=c \geq 1$. Applying the monotonicity together with (1), we get

$$
\begin{aligned}
2 k+c \geq f(2 k)=f(k+k) & \geq f(k)+f(f(k))-1 \\
& \geq f(k)+f(k)-1=2(k+c)-1=2 k+(2 c-1)
\end{aligned}
$$

hence $c \leq 1$ and $f(n) \leq n+1$ for all $n \in \mathbb{N}$. In particular, $f(2007) \leq 2008$.

Now we present a family of examples showing that all values from 1 to 2008 can be realized. Let

$$
f_{j}(n)=\max \{1, n+j-2007\} \quad \text { for } j=1,2, \ldots, 2007 ; \quad f_{2008}(n)= \begin{cases}n, & 2007 \not x n \\ n+1, & 2007 \mid n .\end{cases}
$$

We show that these functions satisfy the condition (1) and clearly $f_{j}(2007)=j$.

To check the condition (1) for the function $f_{j}(j \leq 2007)$, note first that $f_{j}$ is nondecreasing and $f_{j}(n) \leq n$, hence $f_{j}\left(f_{j}(n)\right) \leq f_{j}(n) \leq n$ for all $n \in \mathbb{N}$. Now, if $f_{j}(m)=1$, then the inequality (1) is clear since $f_{j}(m+n) \geq f_{j}(n) \geq f_{j}\left(f_{j}(n)\right)=f_{j}(m)+f_{j}\left(f_{j}(n)\right)-1$. Otherwise,

$$
f_{j}(m)+f_{j}\left(f_{j}(n)\right)-1 \leq(m+j-2007)+n=(m+n)+j-2007=f_{j}(m+n) .
$$

In the case $j=2008$, clearly $n+1 \geq f_{2008}(n) \geq n$ for all $n \in \mathbb{N}$; moreover, $n+1 \geq$ $f_{2008}\left(f_{2008}(n)\right)$ as well. Actually, the latter is trivial if $f_{2008}(n)=n$; otherwise, $f_{2008}(n)=n+1$, which implies $2007 \not h n+1$ and hence $n+1=f_{2008}(n+1)=f_{2008}\left(f_{2008}(n)\right)$.

So, if $2007 \mid m+n$, then

$$
f_{2008}(m+n)=m+n+1=(m+1)+(n+1)-1 \geq f_{2008}(m)+f_{2008}\left(f_{2008}(n)\right)-1 .
$$

Otherwise, $2007 \not \nmid m+n$, hence $2007 \not \nmid m$ or $2007 \not h$. In the former case we have $f_{2008}(m)=m$, while in the latter one $f_{2008}\left(f_{2008}(n)\right)=f_{2008}(n)=n$, providing

$$
f_{2008}(m)+f_{2008}\left(f_{2008}(n)\right)-1 \leq(m+n+1)-1=f_{2008}(m+n) .
$$

Final answer: 1, 2, .., 2008