SAUDI ARABIAN MATHEMATICAL COMPETITIONS

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 $$n-1|a_1+a_2+\cdots +a_{n-1}=\frac{n(n+1)}{2}-a_n=\frac{(n+1)(n-1)}{2}+\frac{n+1}{2}-a_n$$ 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 $$\frac{n+1}{2}-a_n=0\iff a_n=\frac{n+1}{2}$$ It implies that $$a_1+a_2+\cdots +a_{n-1}=\frac{n^2-1}{2}.$$ Continue, we have $$n-2|a_1+a_2+\cdots +a_{n-2}=\frac{n^2-1}{2}-a_{n-1}=\frac{(n-2)(n+1)}{2}+\frac{n+1}{2}-a_{n-1}$$ 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 $n=3$. $\square$