# Answer: 38.

Solution. First, let's provide an example of 38 numbers whose sum is 0 and the sum of their squares is 36. For instance, the numbers $x_{1}=x_{2}=\ldots=x_{19}=\sqrt{\frac{18}{19}}, x_{20}=x_{21}=\ldots=x_{38}=-\sqrt{\frac{18}{19}}$ will work.

Now, we will prove that fewer than 38 numbers will not suffice. Suppose the contrary. Then, among all the numbers, either the positive or the negative ones do not exceed 18. By multiplying all the numbers by -1 if necessary, we can assume that there are no more than 18 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 $y_{1}^{2}+y_{2}^{2}+\ldots+y_{k}^{2}<k \leqslant 18$, and

$$
y_{k+1}^{2}+y_{k+2}^{2}+\ldots+y_{n}^{2} \leqslant y_{k+1}+y_{k+2}+\ldots+y_{n}=-y_{1}-y_{2}-\ldots-y_{k}<k \leqslant 18
$$

Adding the obtained inequalities, we get, $36=y_{1}^{2}+y_{2}^{2}+\ldots+y_{n}^{2}<36$. Contradiction.
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 36?

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