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 40?
Answer: 42.

Solution. First, let's provide an example of 42 numbers whose sum is 0 and the sum of their squares is 40. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{21}=\sqrt{\frac{20}{21}}, x_{22}=x_{23}=\ldots=x_{42}=-\sqrt{\frac{20}{21}}$ will work.

Now, we will prove that fewer than 42 numbers will not suffice. Suppose the contrary. Then, among all the numbers, either the positive or the negative ones do not exceed 20. By multiplying all the numbers by -1 if necessary, we can assume that there are no more than 20 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 $y_{1}^{2}+y_{2}^{2}+\ldots+y_{k}^{2}<k \leqslant 20$, and

$$
y_{k+1}^{2}+y_{k+2}^{2}+\ldots+y_{n}^{2} \leqslant y_{k+1}+y_{k+2}+\ldots+y_{n}=-y_{1}-y_{2}-\ldots-y_{k}<k \leqslant 20
$$

Adding the obtained inequalities, we get, $40=y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}<40$. Contradiction.