Iranian Mathematical Olympiad

We prove by induction that $f(n) = n f(1)$.

Let $c_1, c_2, \dots$, be the sequence of positive integers such that $f(c_i) = p_i^k$. Then one has
$$f(c_i - (d+1)) + f(d+1) \mid f(c_i) = p_i^k$$
and hence $f(c_i - (d+1)) = p_i^j - f(d+1)$ for some positive integer $j \le k$. By pigeonhole principle, there are infinitely many $i$ that gives us the same $j$ and hence we can assume that $j$ is fixed since we can just consider that sequence instead. Similarly, one has $f(c_i - k) = p_i^{j'} - f(k)$ for some fixed positive integer $j'$. Now we have
$$f(1) + f(c_i - (d+1)) \mid f(c_i - d) \implies p_i^j - f(d+1) + f(1) \mid p_i^{j'} - f(d).$$
Let $j' = a j + b$ where $0 \le b < j$. Then our divisibility condition becomes
$$p_i^j - f(d+1) + f(1) \mid (f(d+1) - f(1))^a p_i^b - f(d)$$
but since $b < j$ and $a < k$, the RHS is less than the LHS when $p_i$ is large enough which is a contradiction unless the LHS equals zero. In which case one has
$$(f(d+1) - f(1))^a p_i^b = f(d)$$
and so $b = 0$, giving us $(f(d+1) - f(1))^a = f(d)$. On the other hand, one also has
$$f(1) + f(d) \mid f(d+1)$$
giving us
$$(f(d+1) - f(1))^a + f(1) \mid f(d+1).$$
Letting $f(d+1) - f(1) = c$, one has
$$c^a + f(1) \mid c + f(1)$$
which is impossible unless $c^a = c$, in which case either $c = 1$ or $a = 1$. If $c = 1$, then $f(d) = 1$ and $f(1) + f(d-1) \mid f(d)$ is impossible. Thus it must be that $a = 1$ which gives us $f(d+1) = f(d) + f(1)$. By induction, $f(n) = n f(1)$ as desired.

Now clearly any function satisfying $f(n) = n f(1)$ satisfies the second condition. For the first condition, it is clear that one must have $f(1) = 1$. Hence the only solution is $f(n) = n$ for all $n \in \mathbb{Z}^+$.

$\boxed{f(n) = n}$ for all $n \in \mathbb{Z}^+$.