Determine the largest natural number $m$ such that for each non negative real numbers $a_1 \ge a_2 \ge ... \ge a_{2014} \ge 0$ , it is true that $$\frac{a_1+a_2+...+a_m}{m}\ge \sqrt{\frac{a_1^2+a_2^2+...+a_{2014}^2}{2014}}$$
1. **Restate the problem:**
   We need to determine the largest natural number \( m \) such that for any non-negative real numbers \( a_1 \ge a_2 \ge \dots \ge a_{2014} \ge 0 \), the following inequality holds:
   \[
   \frac{a_1 + a_2 + \dots + a_m}{m} \ge \sqrt{\frac{a_1^2 + a_2^2 + \dots + a_{2014}^2}{2014}}
   \]

2. **Initial observation:**
   To find the largest \( m \), we can test the inequality with specific values. Consider the sequence \( (1, 0, 0, \dots, 0) \) where \( a_1 = 1 \) and \( a_2 = a_3 = \dots = a_{2014} = 0 \). Plugging this into the inequality, we get:
   \[
   \frac{1 + 0 + \dots + 0}{m} \ge \sqrt{\frac{1^2 + 0^2 + \dots + 0^2}{2014}}
   \]
   Simplifying, we have:
   \[
   \frac{1}{m} \ge \sqrt{\frac{1}{2014}}
   \]
   \[
   \frac{1}{m} \ge \frac{1}{\sqrt{2014}}
   \]
   \[
   m \le \sqrt{2014}
   \]
   Since \( \sqrt{2014} \approx 44.9 \), the largest integer \( m \) satisfying this condition is \( m = 44 \).

3. **Verification for \( m = 44 \):**
   We need to show that for \( m = 44 \), the inequality holds for any non-negative sequence \( a_1 \ge a_2 \ge \dots \ge a_{2014} \ge 0 \). The inequality to verify is:
   \[
   \frac{a_1 + a_2 + \dots + a_{44}}{44} \ge \sqrt{\frac{a_1^2 + a_2^2 + \dots + a_{2014}^2}{2014}}
   \]
   Squaring both sides, we get:
   \[
   \left( \frac{a_1 + a_2 + \dots + a_{44}}{44} \right)^2 \ge \frac{a_1^2 + a_2^2 + \dots + a_{2014}^2}{2014}
   \]
   \[
   \frac{(a_1 + a_2 + \dots + a_{44})^2}{44^2} \ge \frac{a_1^2 + a_2^2 + \dots + a_{2014}^2}{2014}
   \]
   \[
   2014(a_1 + a_2 + \dots + a_{44})^2 \ge 1936(a_1^2 + a_2^2 + \dots + a_{2014}^2)
   \]
   This inequality holds because the left-hand side represents the sum of the squares of the first 44 terms, which are the largest terms in the sequence, and the right-hand side represents the sum of the squares of all 2014 terms. Since \( a_1 \ge a_2 \ge \dots \ge a_{2014} \ge 0 \), the sum of the squares of the first 44 terms will be larger or equal to the sum of the squares of the remaining terms.

Conclusion:
\[
\boxed{m = 44}
\]