China Mathematical Olympiad

When $n=1$, $\cos\theta_1 = (1+\tan^2\theta_1)^{-\frac{1}{2}} = \frac{\sqrt{3}}{3}$. Hence $\lambda = \frac{\sqrt{3}}{3}$.

When $n=2$, we can prove
$$\cos \theta_1 + \cos \theta_2 \le \frac{2\sqrt{3}}{3}, \qquad \textcircled{1}$$
and when $\theta_1 = \theta_2 = \arctan\sqrt{2}$, the equality holds.
In fact,
$$\textcircled{1} \Leftrightarrow \cos^2\theta_1 + \cos^2\theta_2 + 2\cos\theta_1 \cdot \cos\theta_2 \le \frac{3}{4},$$
$$\text{that is, } \frac{1}{1+\tan^2\theta_1} + \frac{1}{1+\tan^2\theta_2} + 2\sqrt{\frac{1}{(1+\tan^2\theta_1)(1+\tan^2\theta_2)}} \le \frac{3}{4}. \qquad \textcircled{2}$$
From $\tan \theta_1 \cdot \tan \theta_2 = 2$, we get
$$\textcircled{2} \Leftrightarrow \frac{2+\tan^2\theta_1 + \tan^2\theta_2}{5+\tan^2\theta_1 + \tan^2\theta_2} + 2\sqrt{\frac{1}{5+\tan^2\theta_1 + \tan^2\theta_2}} \le \frac{3}{4}. \qquad \textcircled{3}$$
Write $x = \tan^2\theta_1 + \tan^2\theta_2$, then
$$\textcircled{3} \Leftrightarrow 2\sqrt{\frac{1}{5+x}} \le \frac{14+x}{3(5+x)},$$
$$\text{that is,}\qquad 36(5+x) \le 196+28x+x^2. \qquad \textcircled{4}$$
$$\text{Obviously,}\qquad \textcircled{4} \Leftrightarrow x-8x+16 = (x-4)^2 \ge 0.$$
$$\text{Hence, } \lambda = \frac{2\sqrt{3}}{3}.$$

When $n \ge 3$, there is no loss of generality in supposing $\theta_1 \ge \theta_2 \ge \dots \ge \theta_n$, then
$$\tan \theta_1 \cdot \tan \theta_2 \cdot \tan \theta_3 \ge 2\sqrt{2}.$$
Since $\cos \theta_i = \sqrt{1 - \sin^2 \theta_i} < 1 - \frac{1}{2} \sin^2 \theta_i$, so
$$\cos \theta_2 + \cos \theta_3 < 2 - \frac{1}{2}(\sin^2 \theta_2 + \sin^2 \theta_3) < 2 - \sin \theta_2 \cdot \sin \theta_3.$$
From $\tan^2 \theta_1 \ge \frac{8}{\tan^2 \theta_2 \cdot \tan^2 \theta_3}$, we have
$$\frac{1}{\cos^2 \theta_1} \ge \frac{8 + \tan^2 \theta_2 \cdot \tan^2 \theta_3}{\tan^2 \theta_2 \cdot \tan^2 \theta_3},$$
that is,
$$\begin{aligned} \cos \theta_1 &\le \frac{\tan \theta_2 \cdot \tan \theta_3}{\sqrt{8 + \tan^2 \theta_2 \cdot \tan^2 \theta_3}} \\ &= \frac{\sin \theta_2 \cdot \sin \theta_3}{\sqrt{8\cos^2 \theta_2 \cdot \cos^2 \theta_3 + \sin^2 \theta_2 \cdot \sin^2 \theta_3}}. \end{aligned}$$
Hence
$$\cos \theta_2 + \cos \theta_3 + \cos \theta_1$$
$$< 2 - \sin \theta_2 \cdot \sin \theta_3 \left[ 1 - \frac{1}{\sqrt{8\cos^2 \theta_2 \cdot \cos^2 \theta_3 + \sin^2 \theta_2 \cdot \sin^2 \theta_3}} \right].$$
Note that
$$8\cos^2 \theta_2 \cdot \cos^2 \theta_3 + \sin^2 \theta_2 \cdot \sin^2 \theta_3 \ge 1$$
$$\begin{aligned} \Leftrightarrow & 8 + \tan^2 \theta_2 \tan^2 \theta_3 \ge \frac{1}{\cos^2 \theta_2 \cdot \cos^2 \theta_3} \\ &= (1 + \tan^2 \theta_2)(1 + \tan^2 \theta_3) \\ \Leftrightarrow & \tan^2 \theta_2 + \tan^2 \theta_3 \le 7. \end{aligned}$$
If this does not hold, then $\tan^2 \theta_2 + \tan^2 \theta_3 > 7$,
$$\tan^2 \theta_1 \ge \tan^2 \theta_2 > \frac{7}{2}$$
so
$$\cos \theta_1 \le \cos \theta_2 < \sqrt{\frac{1}{1+\frac{7}{2}}} = \frac{\sqrt{2}}{3}.$$
Thus
$$\cos \theta_1 + \cos \theta_2 + \cos \theta_3 < \frac{2\sqrt{2}}{3} + 1 < 2,$$
that is, the bound holds too.

$$\cos \theta_1 + \cos \theta_2 + \cos \theta_3 + \dots + \cos \theta_n < n - 1.$$
On the other hand, if we take $\theta_2 = \theta_3 = \dots = \theta_n = \alpha > 0$, $\alpha \to 0$, then
$$\theta_1 = \arctan \frac{2^n}{(\tan \alpha)^{n-1}}.$$
Obviously, $\theta_1 \to \frac{\pi}{2}$, thus
$$\cos \theta_1 + \cos \theta_2 + \cos \theta_3 + \dots + \cos \theta_n \to n - 1.$$
Consequently, we get $\lambda = n - 1$.