Problem 1497

The smallest possible value of $n$ is $42$.
Firstly, it is obvious that $40 = a_1^2 + a_2^2 + \cdots + a_n^2 < 1 + 1 + \cdots + 1 = n$. Suppose $n = 41$. WLOG assume $a_1 \le a_2 \le \cdots \le a_n$ and $a_j \ne 0$ for each $j$. Note that $a_1 < 0 < a_{41}$ since $a_1 + a_2 + \cdots + a_{41} = 0$. Let $k$ be the index such that $a_k < 0 < a_{k+1}$. By flipping the signs of all terms if necessary, we may assume $k \le \lfloor \frac{41}{2} \rfloor = 20$. We have
$$\begin{aligned} 40 &= a_1^2 + a_2^2 + \cdots + a_{41}^2 \\ &= (a_1^2 + a_2^2 + \cdots + a_k^2) + (a_{k+1}^2 + \cdots + a_{41}^2) \\ &< (a_1^2 + a_2^2 + \cdots + a_k^2) + (a_{k+1} + \cdots + a_{41}) \\ &= (a_1^2 + a_2^2 + \cdots + a_k^2) - (a_1 + a_2 + \cdots + a_k) \\ &< 2k. \end{aligned}$$
This contradicts with the assumption $k \le 20$. Therefore, $n \ge 42$. It remains to construct such a sequence when $n = 42$. Indeed, we can take $a_j = -\sqrt{\frac{20}{21}}$ for $1 \le j \le 21$ and $a_j = \sqrt{\frac{20}{21}}$ for $22 \le j \le 42$.