Maybe the problem is missing something, but we can do better than $1000$. Denote $V_k =a_1^2+a_2^2+ \ldots+a_k^2$ We have : \[\frac{1}{V_k} \ge \sum_{i=k}^{n-1}\left(\frac{1}{V_i} - \frac{1}{V_{i+1}} \right)=\sum_{i=k}^{n-1}\frac{a_{i+1}^2}{V_iV_{i+1}} \ge \sum_{i=k}^{n-1}\frac{a_{i+1}^2}{V_{i+1}^2} \ge \sum_{i=k}^{n-1} \frac{1}{n^2} = \frac{n-k}{n^2} \]For $k= \frac{9n}{10}$ we get $\sum_{i=1}^{\frac{9n}{10}} a_i^2 \le 10n$. If we have no more than $ \frac{n}{10}$ that are $\le\frac{10}{\sqrt{8}}$ then $\sum_{i=1}^{\frac{9n}{10}} a_i^2 >\frac{8}{10}n\left(\frac{10}{\sqrt{8}}\right)^2=10n$ which is a contradiction. So we have at least $ \frac{n}{10}$ indices $k$ such that $a_k\le\frac{10}{\sqrt{8}}$

Abdek wrote: Maybe the problem is missing something, but we can do better than $1000$. Denote $V_k =a_1^2+a_2^2+ \ldots+a_k^2$ We have : \[\frac{1}{V_k} \ge \sum_{i=k}^{n-1}\left(\frac{1}{V_i} - \frac{1}{V_{i+1}} \right)=\sum_{i=k}^{n-1}\frac{a_{i+1}^2}{V_iV_{i+1}} \ge \sum_{i=k}^{n-1}\frac{a_{i+1}^2}{V_{i+1}^2} \ge \sum_{i=k}^{n-1} \frac{1}{n^2} = \frac{n-k}{n^2} \]For $k= \frac{9n}{10}$ we get $\sum_{i=1}^{\frac{9n}{10}} a_i^2 \le 10n$. If we have no more than $ \frac{n}{10}$ that are $\le\frac{10}{\sqrt{8}}$ then $\sum_{i=1}^{\frac{9n}{10}} a_i^2 >\frac{8}{10}n\left(\frac{10}{\sqrt{8}}\right)^2=10n$ which is a contradiction. So we have at least $ \frac{n}{10}$ indices $k$ such that $a_k\le\frac{10}{\sqrt{8}}$ yes ,it is very corect!