Obviously $n>40$. I am to claim that $n=41$ doesn't work. Suppose contrary, WLOG assume $\sum\limits_{i=1}^{m}a_i=-\left(\sum\limits_{i=m+1}^{41} a_i\right)=k$. Since either $m\le 20$ or $41-m\le 20$ and $\mid a_i\mid <1$, it follows that $k< 20$. Lemma. Let $a_1+a_2+\cdots+a_r=S$ where $a_i(i=1,2,\cdots ,r)$ are both positive or negative, denote $f(a_1,a_2,\cdots,a_r,S)=x_1^2+\cdots+x_n^2$. If $\mid a_1+a_2\mid \ge 1$, then \[f(a_1,a_2,\cdots,a_r,S)< f(1,a_1+a_2-1,\cdots,a_r,S).\]If $\mid a_1+a_2\mid <1$, then replace $(1,a_1+a_2-1)$ with $(a_1+a_2,0)$. Proof of the lemma. It suffices to prove that \[a_1^2+a_2^2\le 1+(a_1+a_2-1)^2,\]which is equivalent to $2(a_1-1)(a_2-1)\ge 0$.$\square$ Back to the main problem, WLOG assume $m\ge 21$, so $41-m\le 20 \implies a_{m+1}^2+\cdots+a_{41}^2<20$. From the above lemma we get \[f(a_1,a_2,\cdots,a_m,k)< f(1,a_1+a_2-1,\cdots,a_m,k)< f(\underbrace{1,1,\cdots,1}_{\left[k\right] \text{terms}},k-\left[k\right],0,\cdots,k)< f(\underbrace{1,1,\cdots,1}_{\left[k\right] \text{terms}},1,0,\cdots,k)=\left[k\right]+1\le 20,\]implying that $40=\sum\limits_{i=1}^{41}a_i^2<20+20=40$, contradiction. So we conclude that $n\ge 42$. Let $a_1=a_2=\cdots =a_{21}=\sqrt{\frac{20}{21}}$, $a_{21}=a_{22}=\cdots=a_{42}=-\sqrt{\frac{20}{21}}$. It's easy to check they satisfy the condition. So the answer is $\boxed{n=42}$.

Do we have any file which invole with this problem ?

why $k>20.24$ ?

Clearly $n>40$, we now show $n=41$ does not work. WLOG assume $a_1,a_2,\ldots ,a_k$ are negative, $a_{k+1},a_{k+2},\ldots ,a_{41}$ are positive, and $k\le 20$. Then $|a_1+a_2+\ldots +a_k|=a_{k+1}+a_{k+2}+\ldots +a_{41}<20$. We have $a_1^2+a_2^2+\ldots + a_k^2<20$. Note that $a_i^2<|a_i|$, hence $a_{k+1}^2+a_{k+2}^2+\ldots +a_{41}^2<a_{k+1}+a_{k+2}+\ldots +a_{41}<20$, so $a_1^2+a_2^2+\ldots +a_{41}^2<40\Rightarrow n>41$. $n=42$ works from the construction shown above, hence $n=42$.

leeky wrote: Clearly $n>40$, we now show $n=41$ does not work. WLOG assume $a_1,a_2,\ldots ,a_k$ are negative, $a_{k+1},a_{k+2},\ldots ,a_{41}$ are positive, and $k\le 20$. Then $|a_1+a_2+\ldots +a_k|=a_{k+1}+a_{k+2}+\ldots +a_{41}<20$. We have $a_1^2+a_2^2+\ldots + a_k^2<20$. Note that $a_i^2<|a_i|$, hence $a_{k+1}^2+a_{k+2}^2+\ldots +a_{41}^2<a_{k+1}+a_{k+2}+\ldots +a_{41}<20$, so $a_1^2+a_2^2+\ldots +a_{41}^2<40\Rightarrow n>41$. $n=42$ works from the construction shown above, hence $n=42$. Nice Just use $a_i^2<\mid a_i\mid$ and my solution is quite complicated.

It's really just a rip-off of Iran problem, I guess it was Iran's second round problem(but for the sum=20, which led to the answer 22)

I wonder if there exists an official website where I can download the past problems and solutions.