Problem 2. For what least $n$ do there exist $n$ numbers from the interval $(-1 ; 1)$ such that their sum is 0 and the sum of their squares is 30?
Answer: 32.

Solution. First, let's provide an example of 32 numbers whose sum is 0 and the sum of their squares is 30. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{16}=\sqrt{\frac{15}{16}}, x_{17}=x_{18}=\ldots=x_{32}=-\sqrt{\frac{15}{16}}$ will work.

Now, we will prove that fewer than 32 numbers will not suffice. Suppose the contrary. Then, among all the numbers, either positive or negative, there are no more than 15. By multiplying all the numbers by -1 if necessary, we can assume that there are no more than 15 negative numbers.

Let $y_{1}, y_{2}, \ldots, y_{k}$ be all the negative numbers, and $y_{k+1}, \ldots, y_{n}$ be all the non-negative numbers. Then $y_{1}^{2}+y_{2}^{2}+\ldots+y_{k}^{2}<k \leqslant 15$, and

$$
y_{k+1}^{2}+y_{k+2}^{2}+\ldots+y_{n}^{2} \leqslant y_{k+1}+y_{k+2}+\ldots+y_{n}=-y_{1}-y_{2}-\ldots-y_{k}<k \leqslant 15
$$

Adding the obtained inequalities, we get, $30=y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}<30$. Contradiction.