Answer: $f(n) = 0, \forall n \in N$

If $f(1) \neq 0$, then take $m = 1, n = |f(1)| + 1$, since $1|\sum_{d|n} f(d)$, so $n|f(1)$ but $0 < |f(1)| < n$, a contradiction. Hence $f(1) = 0$.

Below we prove that for all positive integers $k$, $f(2^k) = 0$.
By contradiction, suppose there exists a positive integer $k$ such that $f(2^k) \neq 0$, and suppose $t$ is the smallest such one, then for all $m > |f(2^t)|$ we have
$m|\sum_{d|2^{t-1}} f(d)$ but $m$ does not divide $\sum_{d|2^t} f(d)$,
that is, $2^{t-1}|f(m)$ but $2^t$ does not divide $f(m)$.
Take a prime $p > |f(2^t)|$, and substitute $n = p^2, m = 2^t$, we get
$$2^t|(f(1)+f(p)+f(p^2)),$$
so $p^2|f(2^t)$, but $0 < |f(2^t)| < p < p^2$, a contradiction.
Therefore $f(2^k) = 0, \forall k \in N$.

For any positive integer $m$, take a sufficiently large $n = 2^k$, then
$$m|(f(1)+\cdots+f(2^k)) \Rightarrow 2^k|f(m),$$
since $k$ can be arbitrarily large, therefore only $f(m) = 0$. Q.E.D.!