Find the smallest positive real number $k$ such that the following inequality holds $$\left|z_{1}+\ldots+z_{n}\right| \geqslant \frac{1}{k}\big(\left|z_{1}\right|+\ldots+\left|z_{n}\right|\big) .$$ for every positive integer $n \geqslant 2$ and every choice $z_{1}, \ldots, z_{n}$ of complex numbers with non-negative real and imaginary parts. 

[Hint: First find $k$ that works for $n=2$. Then show that the same $k$ works for any $n \geqslant 2$.]
1. **Expressing the complex numbers in terms of their real and imaginary parts:**
   Let \( z_i = a_i + i b_i \) for \( 1 \leq i \leq n \), where \( a_i \) and \( b_i \) are non-negative real numbers. 

2. **Summing the real and imaginary parts:**
   Define \( \alpha = a_1 + a_2 + \cdots + a_n \) and \( \beta = b_1 + b_2 + \cdots + b_n \).

3. **Magnitude of the sum of complex numbers:**
   The magnitude of the sum of the complex numbers is given by:
   \[
   |z_1 + z_2 + \cdots + z_n| = \left| \sum_{i=1}^n z_i \right| = \sqrt{\alpha^2 + \beta^2}
   \]

4. **Sum of the magnitudes of the complex numbers:**
   The sum of the magnitudes of the individual complex numbers is:
   \[
   |z_1| + |z_2| + \cdots + |z_n| = \sum_{i=1}^n |z_i| = \sum_{i=1}^n \sqrt{a_i^2 + b_i^2}
   \]
   Since \( a_i \) and \( b_i \) are non-negative, we have:
   \[
   |z_i| = \sqrt{a_i^2 + b_i^2} \leq a_i + b_i
   \]
   Therefore:
   \[
   |z_1| + |z_2| + \cdots + |z_n| \leq (a_1 + b_1) + (a_2 + b_2) + \cdots + (a_n + b_n) = \alpha + \beta
   \]

5. **Applying the Cauchy-Schwarz inequality:**
   By the Cauchy-Schwarz inequality in \(\mathbb{R}^2\), we have:
   \[
   \sqrt{\alpha^2 + \beta^2} \geq \frac{\alpha + \beta}{\sqrt{2}}
   \]

6. **Finding the smallest \( k \):**
   We need to find the smallest positive real number \( k \) such that:
   \[
   \left| z_1 + z_2 + \cdots + z_n \right| \geq \frac{1}{k} \left( |z_1| + |z_2| + \cdots + |z_n| \right)
   \]
   Using the inequality from the Cauchy-Schwarz inequality, we get:
   \[
   \sqrt{\alpha^2 + \beta^2} \geq \frac{\alpha + \beta}{\sqrt{2}}
   \]
   Therefore, we can set:
   \[
   \frac{1}{k} = \frac{1}{\sqrt{2}} \implies k = \sqrt{2}
   \]

7. **Verification for \( n = 2 \):**
   For \( n = 2 \), let \( z_1 = a_1 + i b_1 \) and \( z_2 = a_2 + i b_2 \). Then:
   \[
   |z_1 + z_2| = \sqrt{(a_1 + a_2)^2 + (b_1 + b_2)^2}
   \]
   and:
   \[
   |z_1| + |z_2| \leq (a_1 + b_1) + (a_2 + b_2) = (a_1 + a_2) + (b_1 + b_2)
   \]
   By the Cauchy-Schwarz inequality:
   \[
   \sqrt{(a_1 + a_2)^2 + (b_1 + b_2)^2} \geq \frac{(a_1 + a_2) + (b_1 + b_2)}{\sqrt{2}}
   \]
   Thus, \( k = \sqrt{2} \) works for \( n = 2 \).

8. **Generalization for \( n \geq 2 \):**
   The same argument holds for any \( n \geq 2 \) by summing the real and imaginary parts and applying the Cauchy-Schwarz inequality.

\(\blacksquare\)

The final answer is \( \boxed{ \sqrt{2} } \)