Suppose that $a_1, a_2, \dots, a_n$ satisfies the conditions. First, we have
$$20 = a_1^2 + a_2^2 + \dots + a_n^2 < \underbrace{1 + 1 + \dots + 1}_{n} = n.$$
So $21 \le n$. We want to show that $n = 22$ is the answer. So we prove that there are not $21$ numbers $a_1, a_2, \dots, a_{21}$ in the interval $(-1, 1)$ such that $a_1 + a_2 + \dots + a_{21} = 0$ and $a_1^2 + a_2^2 + \dots + a_{21}^2 = 20$. Assume that sequence $a_i$ is in increasing order so $a_1 \le \frac{a_1 + a_2 + \dots + a_{21}}{21} \le a_{21}$. Thus $a_1 \le 0 \le a_{21}$. But because of minimality of number $21$, we have $a_i \ne 0$ for $1 \le i \le 21$. So there exist a unique number $1 \le k < 21$ such that
$$-1 < a_1 \le a_2 \le \dots \le a_k < 0 < a_{k+1} \le \dots \le a_{21} < 1.$$
We know that numbers $-a_1, -a_2, \dots, -a_{21}$ satisfies the problem condition, too.
Thereby we can assume that $k \le \frac{21}{2}$ and since $k \in \mathbb{Z}$ we have $k \le 10$.
Now for every $k+1 \le i \le 21$. We have $0 < a_i < 1$, so $0 < a_i^2 < a_i$.
$$\begin{aligned} 20 &= a_1^2 + a_2^2 + \dots + a_{21}^2 = (a_1^2 + \dots + a_k^2) + (a_{k+1}^2 + \dots + a_{21}^2) \\ &< (a_1^2 + \dots + a_k^2) + (a_{k+1} + \dots + a_{21}) \\ &< (a_1^2 + \dots + a_k^2) + (-a_1 - a_2 - \dots - a_k) \\ &< 2k \le 20. \end{aligned}$$

This contradiction shows that $n \ge 22$. The following numbers are an example for $n = 22$ and so the answer is $22$.
$$a_i = \sqrt{\frac{11}{10}} \quad (1 \le i \le 11) \quad \text{and} \quad a_i = -\sqrt{\frac{11}{10}} \quad (12 \le i \le 22). \quad \square$$