For the given problem the minimal number of elements is given by $3$ numbers $$ (41,42,110) \text{ and } (20,77,96) $$ Is an example.
The set of positive numbers with the smallest sum is instead given by $$ (1,5,6) \text{ and } (2,3,7) $$
This is the case with $k=2$ of the Prouhet–Tarry–Escott problem
It is known for, that for $2 \le k \le 9$ and $k=11$ we can find two sets with $n=k+1$ numbers that satisfy the equation. In general there is always a solution in which $n=2^k$ (see link for details)
Update: As said in the comments is actually sufficient to consider $n=\frac{1}{2}k(k+1)+1$ (see references in the comments)