, 2016

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,
$$\begin{equation*} f(p) \leqslant \frac{1}{3}\left(p^{2}+2 p-2\right) \tag{2} \end{equation*}$$
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.