[Number Theory]
Problem: Denote by $\mathbb{N}$ the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for all positive integers $m$ and $n$, the integer $f(m)+f(n)-m n$ is nonzero and divides $m f(m)+n f(n)$.

Solution: It is given that

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

Taking $m=n=1$ in (1), we have $2 f(1)-1 \mid 2 f(1)$. Then $2 f(1)-1 \mid 2 f(1)-(2 f(1)-1)=1$ and hence $f(1)=1$.

Let $p \geqslant 7$ be a prime. Taking $m=p$ and $n=1$ in (1), we have $f(p)-p+1 \mid p f(p)+1$ and hence

$$
f(p)-p+1 \mid p f(p)+1-p(f(p)-p+1)=p^{2}-p+1 .
$$

If $f(p)-p+1=p^{2}-p+1$, then $f(p)=p^{2}$. If $f(p)-p+1 \neq p^{2}-p+1$, as $p^{2}-p+1$ is an odd positive integer, we have $p^{2}-p+1 \geqslant 3(f(p)-p+1)$, that is,

$$
f(p) \leqslant \frac{1}{3}\left(p^{2}+2 p-2\right)
\tag{2}
$$

Taking $m=n=p$ in (1), we have $2 f(p)-p^{2} \mid 2 p f(p)$. This implies

$$
2 f(p)-p^{2} \mid 2 p f(p)-p\left(2 f(p)-p^{2}\right)=p^{3} .
$$

By $(2)$ and $f(p) \geqslant 1$, we get

$$
-p^{2}<2 f(p)-p^{2} \leqslant \frac{2}{3}\left(p^{2}+2 p-2\right)-p^{2}<-p
$$

since $p \geqslant 7$. This contradicts the fact that $2 f(p)-p^{2}$ is a factor of $p^{3}$. Thus we have proved that $f(p)=p^{2}$ for all primes $p \geqslant 7$.

Let $n$ be a fixed positive integer. Choose a sufficiently large prime $p$. Consider $m=p$ in (1). We obtain

$$
f(p)+f(n)-p n \mid p f(p)+n f(n)-n(f(p)+f(n)-p n)=p f(p)-n f(p)+p n^{2} .
$$

As $f(p)=p^{2}$, this implies $p^{2}-p n+f(n) \mid p\left(p^{2}-p n+n^{2}\right)$. As $p$ is sufficiently large and $n$ is fixed, $p$ cannot divide $f(n)$, and so $\left(p, p^{2}-p n+f(n)\right)=1$. It follows that $p^{2}-p n+f(n) \mid p^{2}-p n+n^{2}$ and hence

$$
p^{2}-p n+f(n) \mid\left(p^{2}-p n+n^{2}\right)-\left(p^{2}-p n+f(n)\right)=n^{2}-f(n) .
$$

Note that $n^{2}-f(n)$ is fixed while $p^{2}-p n+f(n)$ is chosen to be sufficiently large. Therefore, we must have $n^{2}-f(n)=0$ so that $f(n)=n^{2}$ for any positive integer $n$.

Finally, we check that when $f(n)=n^{2}$ for any positive integer $n$, then

$$
f(m)+f(n)-m n=m^{2}+n^{2}-m n
$$

and

$$
m f(m)+n f(n)=m^{3}+n^{3}=(m+n)\left(m^{2}+n^{2}-m n\right) .
$$

The latter expression is divisible by the former for any positive integers $m, n$. This shows $f(n)=n^{2}$ is the only solution.

Final answer: $f(n)=n^{2}$