These are exactly the prime numbers!

Indeed, if $n=ab$, we can take $a_{1}=0$ and $a_{2}=\cdots=a_{n}=a$. The sum of the $a_{i}$ is $a(n-1)$, which is not divisible by $n$. However, for any $1 \leqslant i \leqslant n$, if $i+b-1 \leqslant n$, then the number $a_{i}+\cdots+a_{i+b-1}=ab=n$ is divisible by $n$. If $i+b-1>n$, then the number $a_{i}+\cdots+a_{i+b}=ab=n$ is divisible by $n$.
Conversely, suppose $n$ is prime, and let $a_{1}, \ldots, a_{n}$ be integers whose sum is not divisible by $n$. If $n$ does not satisfy the property, then for any index $i$, there exists $j(i)$ with $i+1 \leqslant j(i) \leqslant i+n$ such that

$$a_{i}+a_{i+1}+\cdots+a_{j(i)-1}$$

is divisible by $n$. Furthermore, since the sum of the $a_{i}$ is not divisible by $n$, we cannot have $j(i)=i+n$, so $i+1 \leqslant j(i) \leqslant i+n-1$. We then define by induction a sequence of indices $\left(i_{n}\right)$ by $i_{1}=1$ and $i_{n+1}=j\left(i_{n}\right)$. We know that for all $k$, the integer

$$a_{i_{k}}+\cdots+a_{i_{k+1}-1}$$

is divisible by $n$; therefore, by summing, for all indices $k<\ell$, the integer

$$a_{i_{k}}+\cdots+a_{i_{\ell}-1}$$

is divisible by $n$. By the pigeonhole principle, there exist $1 \leqslant k<\ell \leqslant n+1$ such that $i_{k} \equiv i_{\ell}(\bmod n)$. The number of terms in the sum (1) is then $i_{\ell}-i_{k}$, which is divisible by $n$, so each of the $a_{i}$ appears exactly $\frac{i_{\ell}-i_{k}}{n}$ times. Moreover, we know that $i_{j+1}-i_{j} \leqslant n-1$ for all $j$ and that $\ell-k \leqslant n$, so $i_{\ell}-i_{k} \leqslant n(n-1)$. The sum (1) is therefore

$$\frac{i_{\ell}-i_{k}}{n} \times \sum_{i=1}^{n} a_{i} .$$

But $\frac{i_{\ell}-i_{k}}{n} \leqslant n-1$ and cannot be divisible by $n$, and the sum of the $a_{i}$ is not divisible by $n$ either. Since $n$ is prime, the sum (1) is not divisible by $n$, leading to a contradiction.