Clearly the zero function satisfies the condition. Let us prove that this is the only one. In the end we will do an induction (or rather infinite descent) but for the sake of illustration let us show the first step in full detail. Putting $m=1$ one finds that $f(1)=0$. Hence for a prime $p$ we have $p \mid f(n) \Leftrightarrow n \mid f(p) \Leftrightarrow p \mid \sum_{d \mid n} f(d)$. In particular, if we choose $p=2$ and $n=q^2$ for a prime $q$, we conclude that \[2 \mid f(q^2) \Leftrightarrow 2 \mid f(q)+f(q^2).\]In particular $2 \mid f(q)$ for all primes $q$. Hence also $q \mid f(2)$ and hence $f(2)=0$ and hence $2 \mid f(m)$ for all $m$. Now one only needs to observe that one can repeat this step with higher powers of $2$, eventually leading to $2^k \mid f(m)$ for all $m,k$ and hence $f(m)=0$ for all $m$. Detailed argumentSuppose that $f(m) \ne 0$ for some $m$. Then there is some $k$ such that $2^{k-1} \mid f(n)$ for all $n$ but $2^k \nmid f(m_0)$ for some special $m_0$. (We have seen above that $k \ge 2$ but this is not important now.) Choosing $n=2^{i}$ for $i=1,2,\dots,k-1$, this easily implies that $f(2)=f(4)=\dots=f(2^{k-1})=0$. Choosing $n=2^k$ we find that \[2^k \mid f(m) \Leftrightarrow m \mid f(2^k) \Leftrightarrow 2^k \mid \sum_{d \mid m} f(d).\]Choosing $m=q^2$ for a prime $q$, we find that \[2^k \mid f(q^2) \Leftrightarrow 2^k \mid f(q)+f(q^2).\]But remember that $2^{k-1} \mid f(q), f(q^2)$. So as before this implies that $2^k \mid f(q)$ and hence $q \mid f(2^k)$ for all primes $q$ so that $f(2^k)=0$. But then as before $2^k \mid f(m)$ for all $m$ which is a contradiction to our assumption.