It is easy to see that $n=1$, $n=3$ satisfy the given condition, while $n=2$ does not. We will show that all numbers $n>3$ do not satisfy the condition.

First, we can see that
$$a_{1}+a_{2}+\cdots+a_{n}=1+2+\cdots+n=\frac{n(n+1)}{2}$$
then $\frac{n(n+1)}{2}$ is divisible by $n$, which means $\frac{n+1}{2} \in \mathbb{Z}$ or $n$ is an odd number. We also have
$$n-1 \left\lvert\, a_{1}+a_{2}+\cdots+a_{n-1}=\frac{n(n+1)}{2}-a_{n}=\frac{(n+1)(n-1)}{2}+\frac{n+1}{2}-a_{n}\right.$$
Thus, $\frac{n+1}{2}-a_{n}$ is divisible by $n-1$.

On the other hand,
$$-(n-1)<\frac{n+1}{2}-a_{n}<n-1$$
then we must have
$$\frac{n+1}{2}-a_{n}=0 \Leftrightarrow a_{n}=\frac{n+1}{2}$$
It implies that
$$a_{1}+a_{2}+\cdots+a_{n-1}=\frac{n^{2}-1}{2} .$$
Continue, we have
$$n-2 \left\lvert\, a_{1}+a_{2}+\cdots+a_{n-2}=\frac{n^{2}-1}{2}-a_{n-1}=\frac{(n-2)(n+1)}{2}+\frac{n+1}{2}-a_{n-1}\right.$$
Similarly, we get $a_{n-1}=\frac{n+1}{2}$, then $a_{n-1}=a_{n}$, which is a contradiction.

Hence, the only two satisfied values are $n=1$ and $n=3$. $\square$