Iranian Mathematical Olympiad

Suppose that $a_1,a_2,\dots ,a_n$ satisfies the conditions. First, we have $$20=a_1^2+a_2^2+\cdots +a_n^2<\underset{n}{\underbrace{1+1+\cdots +1}}=n.$$ 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 $a_1+a_2+\cdots +a_{21}=0$ and $a_1^2+a_2^2+\cdots +a_{21}^2=20$. Assume that sequence $a_i$ is in increasing order so $a_1\le \frac{a_1+a_2+\cdots +a_{21}}{21}\le a_{21}$. 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 $$-1<a_1\le a_2\le \cdots \le a_k<0<a_{k+1}\le \cdots \le a_{21}<1.$$ 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 $0<a_i^2<a_i$. $$\begin{array}{rl}20& =a_1^2+a_2^2+\cdots +a_{21}^2=(a_1^2+\cdots +a_k^2)+(a_{k+1}^2+\cdots +a_{21}^2)\\ & <(a_1^2+\cdots +a_k^2)+(a_{k+1}+\cdots +a_{21})\\ & <(a_1^2+\cdots +a_k^2)+(-a_1-a_2-\cdots -a_k)\\ & <2k\le 20.\end{array}$$

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