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 42?
Answer: 44.

Solution. First, let's provide an example of 44 numbers whose sum is 0 and the sum of their squares is 42. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{22}=\sqrt{\frac{21}{22}}, x_{23}=x_{24}=\ldots=x_{44}=-\sqrt{\frac{21}{22}}$ will work.

Now, we will prove that it is impossible to achieve this with fewer than 44 numbers. Suppose the contrary. Then, among all the numbers, either the positive or the negative ones do not exceed 21. By multiplying all the numbers by -1 if necessary, we can assume that there are no more than 21 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 $y_{1}^{2}+y_{2}^{2}+\ldots+y_{k}^{2}<k \leqslant 21$, and

$$
y_{k+1}^{2}+y_{k+2}^{2}+\ldots+y_{n}^{2} \leqslant y_{k+1}+y_{k+2}+\ldots+y_{n}=-y_{1}-y_{2}-\ldots-y_{k}<k \leqslant 21
$$

Adding the obtained inequalities, we get, $42=y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}<42$. Contradiction.