Solution:
If $f(n)=n-1$, then $n$ divides the sum $1+2+\cdots+(n-1)=\frac{(n-1)n}{2}$, which implies that $n$ is odd.

The numbers $n=p^{s}$, where $p>2$ is a prime number and $s \geq 1$, are solutions. Indeed, if $k \in \mathbb{N}$ and $k<p^{s}-1$, then the sum $1+2+\cdots+k=\frac{k(k+1)}{2}$ is not divisible by $p^{s}$ because $k$ and $k+1$ are coprime and less than $p^{s}$.

We shall prove that these are the only solutions. Let $n$ be an odd positive integer which is not a power of a prime number. Then $n=a b$, where $a>1$, $b>1$ and $(a, b)=1$. By the Chinese Remainder Theorem we conclude that there exists an integer $k \in[0, a b-1]$ such that $a \mid k$ and $b \mid k+1$. It is clear that $k \neq 0, k \neq a b-1$ and $a b \left\lvert\, \frac{k(k+1)}{2}\right.$. Therefore $n=a b$ is not a solution.