China Team Selection Test

Find all positive real numbers $\lambda$ such that for all integers $n \geq 2$ and all positive real numbers $a_1, a_2, \ldots, a_n$ with $a_1 + a_2 + \cdots + a_n = n$, the following inequality holds:
$$\sum_{i=1}^n \frac{1}{a_i} - \lambda \prod_{i=1}^{n} \frac{1}{a_i} \leq n - \lambda.$$

To find the values of $\lambda$, we first show that $\lambda < e$ fails. Rewrite the inequality by multiplying both sides by $a_1 a_2 \cdots a_n$:
$$\sum_{i=1}^n a_1 a_2 \cdots a_{i-1} a_{i+1} \cdots a_n - \lambda \leq (n - \lambda) \cdot a_1 a_2 \cdots a_n.$$

Set $a_n = 0$ and $a_1 = a_2 = \cdots = a_{n-1} = \frac{n}{n-1}$. The inequality reduces to:
$$\left(\frac{n}{n-1}\right)^{n-1} - \lambda \leq 0 \implies \lambda \geq \left(\frac{n}{n-1}\right)^{n-1}.$$

Taking the limit as $n \rightarrow \infty$, we get $\lambda \geq e$.

Next, we show that $\lambda \geq e$ satisfies the inequality. Using Lagrange multipliers, the left-hand side of the original inequality is maximized when:
$$-\frac{1}{a_i^2} + \frac{\lambda}{a_i} \cdot \frac{1}{a_1 a_2 \cdots a_n} = \mu \implies \frac{\lambda}{a_1 a_2 \cdots a_n} = \mu a_i + \frac{1}{a_i} \text{ for } i = 1, 2, \ldots, n.$$

Assume $a_i \neq a_j$ for some $i, j$. Then:
$$\mu a_i + \frac{1}{a_i} = \mu a_j + \frac{1}{a_j} \implies \mu = \frac{1}{a_i a_j}.$$

Plugging this in for $\mu$ and using AM-GM, we get:
$$\frac{\lambda}{a_1 a_2 \cdots a_n} = \frac{1}{a_j} + \frac{1}{a_i} \implies \lambda = \frac{a_i + a_j}{a_i a_j} \cdot a_1 a_2 \cdots a_n \leq \left(\frac{a_1 + a_2 + \cdots + a_n}{n-1}\right)^{n-1} = \left(\frac{n}{n-1}\right)^{n-1} < e,$$
which is a contradiction. Hence, $a_i = a_j$ for all $i, j$.

The inequality is easy to verify when all $a_i$ are equal. For boundary cases, set $a_n = 0$ and consider the rewritten form of the inequality. It suffices to show that $a_1 a_2 \cdots a_{n-1} \leq \lambda$. By AM-GM:
$$a_1 a_2 \cdots a_{n-1} \leq \left(\frac{n}{n-1}\right)^{n-1} \leq e \leq \lambda.$$

Thus, the values of $\lambda$ that satisfy the inequality are:
$$\lambda \geq e.$$

The answer is: $\boxed{\lambda \geq e}$.