Problem 1501

1. Let $a_1, a_2, \dots, a_n$ be the given non-zero integers, and let $S = \sum_{i=1}^n a_i$ be the sum of these integers.
2. According to the problem, each $a_i$ is divisible by the sum of the other $n-1$ numbers. This can be written as:
$$S - a_i \mid a_i \quad \text{for all } i.$$
3. This implies that $S - a_i \mid S$ for all $i$. Let $k_i$ be an integer such that:
$$k_i = \frac{S}{S - a_i}.$$
4. Rearranging the equation, we get:
$$a_i = \frac{S \cdot (k_i - 1)}{k_i}.$$
5. Summing up all $a_i$, we have:
$$S = \sum_{i=1}^n a_i = \sum_{i=1}^n \frac{S \cdot (k_i - 1)}{k_i}.$$
6. Factoring out $S$ from the sum, we get:
$$S = S \cdot \sum_{i=1}^n \left(1 - \frac{1}{k_i}\right).$$
7. If $S \neq 0$, we can divide both sides by $S$:
$$1 = \sum_{i=1}^n \left(1 - \frac{1}{k_i}\right).$$
8. Simplifying the right-hand side, we get:
$$1 = n - \sum_{i=1}^n \frac{1}{k_i}.$$
9. Rearranging the equation, we obtain:
$$\sum_{i=1}^n \frac{1}{k_i} = n - 1.$$
10. Since $k_i$ are integers, the only possible values for $k_i$ that satisfy this equation are $k_1 = k_2 = \dots = k_{n-2} = 1$ and $k_{n-1} = k_n = 2$. However, this leads to a contradiction because if $k_i = 1$, then $a_i = 0$, which contradicts the given condition that all $a_i$ are non-zero.
11. Therefore, the assumption $S \neq 0$ must be false, implying that $S = 0$.

The final answer is $\boxed{ S = 0 }$.