$f(n) = n^2$.

It is given that
$$(f(m) + f(n) - mn) \mid (mf(m) + nf(n)). \quad (1)$$

1. Take $m = n = 1$ in (1), we have $2f(1) - 1 \mid 2f(1) \Rightarrow f(1) = 1$.

2. Let $p$ be a prime number, and take $(m, n) = (p, 1)$, then we have $f(p) - 1 + p \mid pf(p) + 1$, which is
$$f(p) - p + 1 \mid pf(p) + 1 - p(f(p) - p + 1) = p^2 - p + 1.$$
- If $f(p) - p + 1 = p^2 - p + 1$, then $f(p) = p^2$.
- If $f(p) - p + 1 \neq p^2 - p + 1$, since $p^2 - p + 1$ is odd, we must have $3(f(p) - p + 1) \leq (p^2 - p + 1)$, that is
$$f(p) \leq \frac{1}{3}(p^2 + 2p - 2) \quad (2)$$

Now, take $m = n = p$ in (1), we have $2f(p) - p^2 \mid 2pf(p)$. This implies
$$2f(p) - p^2 \mid 2pf(p) - p(2f(p) - p^2) = p^3.$$
But by (2) and by the fact that $f(p) \geq 1$, we have
$$-p^2 < 2f(p) - p^2 \leq \frac{2}{3}(p^2 + 2p - 2) - p^2 < -p$$
for $p \geq 7$. This contradicts to the fact that $2f(p) - p^2$ is a factor of $p^3$.

Hence, we have proved that $f(p) = p^2$ for all prime $p \ge 7$.

3. Now, fixed $n$, and take prime $p \ge 7$. Set $m = p$ in (2), and we have
$$\begin{aligned} & p^2 + f(n) - pn \\ = & f(p) + f(n) - pn \mid p f(p) + n f(n) - n(f(p) + f(n) - pn) \\ = & p f(p) - n f(p) + pn^2 = p(p^2 - pn + n^2). \end{aligned}$$
However, we can take $p$ sufficiently large such that $p \nmid f(n)$, which means $(p, p^2 - pn + f(n)) = 1$, so $p^2 - pn + f(n) \nmid p^2 - pn + n^2$. Hence,
$$p^2 - pn + f(n) \nmid p^2 - pn + n^2 + (p^2 - pn + f(n)) = n^2 - f(n).$$
Since this must hold for all $p$, but $p^2 - pn + f(n) \to \infty$, the only possibility is $n^2 - f(n) = 0$, that is $f(n) = n^2$.

4. The rest is to check that $f(n) = n^2$ indeed satisfies the conditions.