Croatian Mathematical Olympiad

It is given that
$$f(a) + f(b) - ab \mid af(a) + bf(b). \qquad (3)$$
Taking $a = b = 1$ in (3), we have $2f(1) - 1 \mid 2f(1)$. Then $2f(1) - 1 \mid 2f(1) - (2f(1) - 1) = 1$ and hence $f(1) = 1$.

Let $p \ge 7$ be a prime. Taking $a = p$ and $b = 1$ in (3), we have $f(p) - p + 1 \mid pf(p) + 1$ and hence
$$f(p) - p + 1 \mid pf(p) + 1 - p(f(p) - p + 1) = p^2 - p + 1.$$
If $f(p) - p + 1 = p^2 - p + 1$, then $f(p) = p^2$. If $f(p) - p + 1 \ne p^2 - p + 1$, as $p^2 - p + 1$ is an odd positive integer, we have $p^2 - p + 1 \ge 3(f(p) - p + 1)$, i.e.
$$f(p) \le \frac{1}{3}(p^2 + 2p - 2). \qquad (4)$$
Taking $a = b = p$ in (3), we have $2f(p) - p^2 \mid 2pf(p)$. This implies
$$2f(p) - p^2 \mid 2pf(p) - p(2f(p) - p^2) = p^3.$$
By (4) and $f(p) \ge 1$ we get
$$-p^2 < 2f(p) - p^2 \le \frac{2}{3}(p^2 + 2p - 2) - p^2 < -p,$$
since $p \ge 7$. This contradicts the fact that $2f(p) - p^2$ is a factor of $p^3$. Thus we have proved that $f(p) = p^2$ for all primes $p \ge 7$.

Let $a$ be a fixed positive integer. Choose a sufficiently large prime $p$. Consider $b = p$ in (3). We obtain
$$f(a) + p^2 - pa \mid af(a) + p^3 = a(f(a) + p^2 - pa) + p^3 - p^2a + pa^2,$$
i.e.
$$f(a) + p^2 - pa \mid p(p^2 - pa + a^2).$$
As $p$ is sufficiently large and $a$ is fixed, $p$ cannot divide $f(a)$, and so numbers $f(a) + p^2 - pa$ and $p$ are relatively prime. It follows that
$$f(a) + p^2 - pa \mid p^2 - pa + a^2 = (f(a) + p^2 - pa) + a^2 - f(a),$$
i.e.
$$f(a) + p^2 - pa \mid a^2 - f(a).$$
Note that $a^2 - f(a)$ is fixed while $f(a) + p^2 - pa$ is chosen to be sufficiently large. Therefore, we must have $a^2 - f(a) = 0$, so that $f(a) = a^2$ for any positive integer $a$.

Finally, we check that when $f(a) = a^2$ for any positive integer $a$, then
$$f(a) + f(b) - ab = a^2 + b^2 - ab$$
and
$$af(a) + bf(b) = a^3 + b^3 = (a + b)(a^2 + b^2 - ab).$$
The latter expression is divisible by the former for any positive integers $a$ and $b$. This shows that $f(a) = a^2$ is the only solution.