Croatian Mathematical Olympiad

All such numbers are primes.

Let $n$ be a composite number, i.e. $n = ab$, where $a \ge 2$ and $b \ge 2$ are positive integers.
Observe the sequence
$0, b, b, \dots, b,$
in which the number $b$ is appearing $ab-1$ times. The sum of all numbers in that sequence is $ab^2-b$, which is not divisible by $n=ab$. However, notice that, no matter on the initial index of the sequence, the sum of $a$ (or $a+1$, if we pass through element 0 of the sequence) consecutive numbers in the sequence will be equal to $ab$, and therefore divisible by $n=ab$.

Let $n = p$ be a prime number. Let us assume the opposite: there exist positive integers $a_1, a_2, \dots, a_p$ which all add up to a number not divisible by $p$, but for all $i \in \{1, 2, \dots, p\}$ there exists $j \in \{i+1, i+2, \dots, i+p-1\}$ such that the number $a_i + a_{i+1} + \dots + a_{j-1}$ is divisible by $p$. Reminder: all operations on indices are made using convention $a_i = a_{i-p}$ for $i > p$.

Let us now make a graph consisted of $p$ vertices (indicating numbers $a_1, a_2, \dots, a_p$). The two vertices $a_i$ and $a_j$ are joined with directed edge if the sum $a_i+a_{i+1}+\dots+a_{j-1}$ is divisible by $p$. Due to the assumption, from any vertex there is at least one edge incident to that vertex. Since there are only finitely many vertices in the graph, we conclude that the graph has a cycle.

Consider that cycle and let us sum up all sums which correspond to edges of that cycle. Notice that all numbers $a_1, a_2, \dots, a_p$ appear equally many times in those sums, and let us denote that number of occurrence by $k$. Moreover, in that cycle we have at most $p$ edges, and for each edge the corresponding sum consists of at most $p-1$ addends, since the sum of all numbers in the original sequence is not divisible by $p$. Hence, $k \le p-1$. However, each of those sums along the edges of the cycle is divisible by $p$, so the sum of all those sums in the cycle is $p$ as well, and we have
$$p \mid k(a_1 + a_2 + \dots + a_p).$$
This gives contradiction, since $k < p$, i.e. $p \nmid k$ and $p \nmid a_1 + a_2 + \dots + a_p$.