Solution to the problem as stated: Clearly for $m=2014$ the inequality holds. For $m=2015$ the inequality does not hold, since for any choice of $a_1, a_2, \dots, a_{2014}$, we can make $a_{2015}$ as small as we want to falsify it.

parmenides51 wrote: Determine the largest natural number $m$ such that for each non negative real numbers $a_1 \le a2 \le ... \le a_{2014} \le 0$ , it is true that $$\frac{a_1+a_2+...+a_m}{m}\ge \sqrt{\frac{a_1^2+a_2^2+...+a_{2014}^2}{2014}}$$ Actually the corrected version should be : Determine the largest natural number $m$ such that for each non negative real numbers $a_1 \ge a_2 \ge \dots \ge a_{2014} \ge 0$, then the following holds. $$\frac{a_1 + a_2 + \dots + a_m}{m} \ge \sqrt{\frac{a_1^2 + a_2^2 + \dots + a_{2014}^2}{2014}}$$

I just replaced the incorrect $\le$ with the correct $\ge$, thanks for noticing