Given a positive integer $n$, determine the largest integer $M$ satisfying
$$\lfloor \sqrt{a_1}\rfloor + ... + \lfloor \sqrt{a_n} \rfloor \ge  \lfloor\sqrt{ a_1 + ... + a_n +M \cdot min(a_1,..., a_n)}\rfloor $$
for all non-negative integers $a_1,...., a_n$.

S. Berlov, A. Khrabrov
To determine the largest integer \( M \) satisfying the given inequality for all non-negative integers \( a_1, \ldots, a_n \), we start by analyzing the inequality:

\[
\lfloor \sqrt{a_1} \rfloor + \lfloor \sqrt{a_2} \rfloor + \cdots + \lfloor \sqrt{a_n} \rfloor \ge \lfloor \sqrt{a_1 + a_2 + \cdots + a_n + M \cdot \min(a_1, \ldots, a_n)} \rfloor
\]

1. **Substitute \( a_1 = a_2 = \cdots = a_n = 3 \):**
   \[
   \lfloor \sqrt{3} \rfloor + \lfloor \sqrt{3} \rfloor + \cdots + \lfloor \sqrt{3} \rfloor \ge \lfloor \sqrt{3n + M \cdot 3} \rfloor
   \]
   Since \( \lfloor \sqrt{3} \rfloor = 1 \), the left-hand side becomes:
   \[
   n \cdot 1 = n
   \]
   The inequality now reads:
   \[
   n \ge \lfloor \sqrt{3n + 3M} \rfloor
   \]

2. **Analyze the right-hand side:**
   To satisfy the inequality, we need:
   \[
   n \ge \sqrt{3n + 3M}
   \]
   Squaring both sides, we get:
   \[
   n^2 \ge 3n + 3M
   \]
   Rearranging terms, we obtain:
   \[
   n^2 - 3n \ge 3M
   \]
   Dividing both sides by 3:
   \[
   \frac{n^2 - 3n}{3} \ge M
   \]
   Therefore:
   \[
   M \le \left\lfloor \frac{n^2 - 3n}{3} \right\rfloor
   \]

3. **Conclusion:**
   The largest integer \( M \) that satisfies the inequality for all non-negative integers \( a_1, \ldots, a_n \) is:
   \[
   M = \left\lfloor \frac{n^2 - 3n}{3} \right\rfloor
   \]

The final answer is \( \boxed{\left\lfloor \frac{n^2 - 3n}{3} \right\rfloor} \).