[Combinatorics]
Problem: Determine the greatest positive integer $k$ that satisfies the following property: The set of positive integers can be partitioned into $k$ subsets $A_{1}, A_{2}, \ldots, A_{k}$ such that for all integers $n \geq 15$ and all $i \in\{1,2, \ldots, k\}$ there exist two distinct elements of $A_{i}$ whose sum is $n$.

Solution: There are various examples showing that $k=3$ does indeed have the property under consideration. E.g. one can take

$$
\begin{gathered}
A_{1}=\{1,2,3\} \cup\{3 m \mid m \geq 4\} \\
A_{2}=\{4,5,6\} \cup\{3 m-1 \mid m \geq 4\} \\
A_{3}=\{7,8,9\} \cup\{3 m-2 \mid m \geq 4\}
\end{gathered}
$$

To check that this partition fits, we notice first that the sums of two distinct elements of $A_{i}$ obviously represent all numbers $n \geq 1+12=13$ for $i=1$, all numbers $n \geq 4+11=15$ for $i=2$, and all numbers $n \geq 7+10=17$ for $i=3$. So, we are left to find representations of the numbers 15 and 16 as sums of two distinct elements of $A_{3}$. These are $15=7+8$ and $16=7+9$.

Let us now suppose that for some $k \geq 4$ there exist sets $A_{1}, A_{2}, \ldots, A_{k}$ satisfying the given property. Obviously, the sets $A_{1}, A_{2}, A_{3}, A_{4} \cup \cdots \cup A_{k}$ also satisfy the same property, so one may assume $k=4$.

Put $B_{i}=A_{i} \cap\{1,2, \ldots, 23\}$ for $i=1,2,3,4$. Now for any index $i$ each of the ten numbers $15,16, \ldots, 24$ can be written as sum of two distinct elements of $B_{i}$. Therefore this set needs to contain at least five elements. As we also have $\left|B_{1}\right|+\left|B_{2}\right|+\left|B_{3}\right|+\left|B_{4}\right|=23$, there has to be some index $j$ for which $\left|B_{j}\right|=5$. Let $B_{j}=\left\{x_{1}, x_{2}, x_{3}, x_{4}, x_{5}\right\}$. Finally, now the sums of two distinct elements of $A_{j}$ representing the numbers $15,16, \ldots, 24$ should be exactly all the pairwise sums of the elements of $B_{j}$. Calculating the sum of these numbers in two different ways, we reach

$$
4\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)=15+16+\ldots+24=195
$$

Thus the number 195 should be divisible by 4, which is false. This contradiction completes our solution.
Again we only prove that $k \leq 3$. Assume that $A_{1}, A_{2}, \ldots, A_{k}$ is a partition satisfying the given property. We construct a graph $\mathcal{G}$ on the set $V=\{1,2, \ldots, 18\}$ of vertices as follows. For each $i \in\{1,2, \ldots, k\}$ and each $d \in\{15,16,17,19\}$ we choose one pair of distinct elements $a, b \in A_{i}$ with $a+b=d$, and we draw an $e d g e$ in the $i^{\text {th }}$ color connecting $a$ with $b$. By hypothesis, $\mathcal{G}$ has exactly 4 edges of each color.

Claim. The graph $\mathcal{G}$ contains at most one circuit.

Proof. Note that all the connected components of $\mathcal{G}$ are monochromatic and hence contain at most four edges. Thus also all circuits of $\mathcal{G}$ are monochromatic and have length at most four. Moreover, each component contains at most one circuit since otherwise it should contain at least five edges.

Suppose that there is a 4-cycle in $\mathcal{G}$, say with vertices $a, b, c$, and $d$ in order. Then $\{a+b, b+$ $c, c+d, d+a\}=\{15,16,17,19\}$. Taking sums we get $2(a+b+c+d)=15+16+17+19$ which is impossible for parity reasons. Thus all circuits of $\mathcal{G}$ are triangles.

Now if the vertices $a, b$, and $c$ form such a triangle, then by a similar reasoning the set $\{a+b, b+$ $c, c+a\}$ coincides with either $\{15,16,17\}$, or $\{15,16,19\}$, or $\{16,17,19\}$, or $\{15,17,19\}$. The last of these alternatives can be excluded for parity reasons again, whilst in the first three cases the set $\{a, b, c\}$ appears to be either $\{7,8,9\}$, or $\{6,9,10\}$, or $\{7,9,10\}$, respectively. Thus, a component containing a circuit should contain 9 as a vertex. Therefore there is at most one such component and hence at most one circuit.

By now we know that $\mathcal{G}$ is a graph with $4 k$ edges, at least $k$ components and at most one circuit. Consequently, $\mathcal{G}$ must have at least $4 k+k-1$ vertices. Thus $5 k-1 \leq 18$, and $k \leq 3$.

Final answer: 3