$n\ge22$ Done!

it needs to show that $n\neq 21$ suppose that $n=21$ and $x_1+...x_{21}=0$ and $x_1^2+...x_{21}^2=20$ it's obvious that $x_i\neq 0$ suppose that $x_1,...x_k<0$ and $x_{k+1},...x_{21}>0$ WLOG $k\le 10$ (because we can change the sign of the $x_i$s) $x_1^2+...x_k^2<k \Longrightarrow 20-k< x_{k+1}^2+...x_{21}^2<x_{k+1}+...+x_{21}$ $x_1+...x_k>-k \Longrightarrow x_1+...x_{21}>20-2k\ge 0$ contradiction.

Actually, the problem can be solved by rejecting $n=21$ case by making the variables far from each other...

dashmiz wrote: Actually, the problem can be solved by rejecting $n=21$ case by making the variables far from each other... Yes , it was also my solution .replace $x,y$ positive and negative rsp by $x+a$ , $y-a$ .

@dashmiz

your solutions are exactly right. but I tested some numbers and they proved that it can be 21 real numbers. my numbers are: ten numbers -0.99 nine numbers 0.99 so it results by the supposition of the problem: x+y=0.99 to the sum of them become zero and:x^2+y^2=1.3781 to the sum of the squares become 20 so by solving this two degree equation results: x=0.173 & y=0.8169 so my sugested numbers are: 10*-0.99 , 9*0.99 , 0.173 , 0.8169 so it can be 21 numbers else. if my counting has any problem please tell. thanks.

the numbers you suggest don't have the sum $0$....

Besides http://www.wolframalpha.com/input/?i=19%280.99%29%5E2+%2B%280.173%29%5E2+%2B%280.8169%29%5E2

any solution?

Jjesus wrote: any solution? Of course since $x_i^2 < 1$ for all $i$, $20 = \sum_{i} x_i^2 < n$, implying $n \geq 21$. The case $n = 21$ can be discarded according to post #3. For $n = 22$ there is an example $x_1 = \cdots x_{11} = x$, $x_{12} = \cdots = x_{22} = -x$ where $x = \sqrt{\frac{10}{11}}$.