Auswahlklausur

Solution:

The numbers sought are precisely the primes.

Partial proof 1: No non-prime number satisfies all the conditions.

Let $n=a \cdot b$ with $1<a, b<n$ be a factorization of $n$ into two proper divisors. We choose $m_{i}=a$ for $1 \leq i<n$ as well as $m_{n}=0$. Then the sum $m_{1}+m_{2}+\ldots+m_{n}=(n-1) a$ is obviously not divisible by $n$, since neither factor is divisible by $n$.
Now, for an arbitrary index $i$, we choose the index $j=\left\{\begin{array}{ll}b & \text{ for } 1 \leq i \leq n-b \\ b+1 & \text{ for } n-b<i \leq n\end{array}\right.$ and obtain $m_{i}+m_{i+1}+\ldots+m_{i+j-1}=a \cdot b=n \equiv 0 \bmod n$. With this counterexample the partial proof is complete.

Partial proof 2: Every prime satisfies all the conditions.

Now let $n$ be a prime. For a proof by contradiction we assume that for the numbers $m_{1}, m_{2}, \ldots, m_{n}$, whose sum is not divisible by $n$, there exists for every index $i$ $(1 \leq i \leq n)$ a number $j$ $(1 \leq j \leq n)$ such that the sum $m_{i}+m_{i+1}+\ldots+m_{i+j-1}$ is divisible by $n$. Here we even have $j \neq n$, since the sum of all $m_{i}$ is not divisible by $n$.
Now, for $0 \leq k \leq n-1$, we construct a finite sequence of integers $i_{0}, i_{1}, \ldots, i_{n}$ with $i_{k+1}-i_{k} \leq n-1$ (1), by choosing $m_{i_{k}+1}+m_{i_{k}+2}+\ldots+m_{i_{k+1}} \equiv 0 \bmod n$. The starting index $i_{0}$ is arbitrary, and the new index $i_{k+1}$ is chosen to be the smallest possible one after $i_{k}$ when proceeding cyclically $\bmod n$.
By the pigeonhole principle, among this sequence of $n+1$ indices there exist two distinct numbers $i_{r}$ and $i_{s}$ with $0 \leq r<s \leq n$ that are congruent $\bmod n$. For these we have $\sum_{j=r}^{s-1}\left(m_{i_{j}+1}+m_{i_{j}+2}+\ldots+m_{i_{j+1}}\right) \equiv 0 \bmod n$, since this holds for every bracketed sum.
On the other hand, from $i_{s} \equiv i_{r} \bmod n$ it follows that there exists a positive integer $d$ with $i_{s}-i_{r}=d \cdot n$. By (1), $i_{s}-i_{r} \leq(n-1) n$, so it follows that $d \leq n-1$. Then $\sum_{j=r}^{s-1}\left(m_{i_{j}+1}+m_{i_{j}+2}+\ldots+m_{i_{j+1}}\right)=d\left(m_{1}+m_{2}+\ldots+m_{n}\right)$ cannot be a multiple of $n$, since $n$ is prime and neither $d$ nor $m_{1}+m_{2}+\ldots+m_{n}$ are multiples of $n$ — contradiction! $\square$