China Team Selection Test

Given an integer $n \geq 2$, we aim to find the minimum value of $\lambda$ such that for any real numbers $a_1, a_2, \ldots, a_n$ and $b$, the following inequality holds:
$$\lambda \sum_{i=1}^n \sqrt{|a_i - b|} + \sqrt{n \left| \sum_{i=1}^n a_i \right|} \geq \sum_{i=1}^n \sqrt{|a_i|}.$$

To determine the minimum value of $\lambda$, we use the triangle inequality in the context of square roots. Specifically, we note that for any real numbers $a$ and $b$,
$$\sqrt{|a|} + \sqrt{|b|} \geq \sqrt{|a \pm b|}.$$

Applying this inequality repeatedly, we obtain:
$$\sum_{\substack{i=1 \\ i \neq j}}^n \sqrt{|a_i - b|} + \sqrt{n-1} \sqrt{|a_j - b|} + \sqrt{\left| \sum_{i=1}^n a_i \right|} \geq \sqrt{\left| \sum_{\substack{i=1 \\ i \neq j}}^n a_i - (n-1)b \right|} + \sqrt{|(n-1)b - (n-1)a_j|} + \sqrt{\left| \sum_{i=1}^n a_i \right|}.$$

Simplifying further, we get:
$$\sqrt{\left| \sum_{\substack{i=1 \\ i \neq j}}^n a_i - (n-1)a_j \right|} + \sqrt{\left| \sum_{i=1}^n a_i \right|} \geq \sqrt{n} \sqrt{|a_j|}.$$

Summing over all $j \in \{1, 2, \ldots, n\}$, we derive the desired inequality. Equality holds for $b = a_1 = \cdots = a_{n-1} = -a_n / (n-1) \neq 0$ (and permutations), proving that the optimal constant is:
$$\lambda = \frac{n-1 + \sqrt{n-1}}{\sqrt{n}}.$$

The answer is: $\boxed{\frac{n-1 + \sqrt{n-1}}{\sqrt{n}}}.$