here it is

Let $a_1 , a_2 , \dots , a_n$ be the given numbers and let $S = \sum_{i=1}^n a_i$. We have $S - a_i \mid a_i \implies S - a_i \mid S $ for all $i$. Let $k_i = \frac{S}{S - a_i}$ for $k_i \in \mathbb{Z}$. We have $a_i = \frac{S\cdot (k_i - 1)}{k_i}$. So taking the sum we have, $S = \sum_{i=1}^n a_i = S\cdot \sum_{i=1}^n (1 - \frac{1}{k_i})$. If $S\neq 0$, then we have $\sum_{i=1}^n \frac{1}{k_i} = n-1 \implies k_1 = k_2 = \dots = k_{n-2} = 1$ and $k_{n-1} = k_n = 2$. But this is clearly a contradiction since then $a_1 = a_2 = \dots = a_{n-2} = 0$. So $S = 0$ as desired. $\blacksquare$ @below $10 = 6 + 2 + 2 \nmid 2$

$(6,2,2,2)?$

@above oooh ok i see, thanks

nice problem but easy to P3