16. Given a positive integer $n$, find the smallest positive number $\lambda$, such that for any $\theta_{i}^{-} \in\left(\theta, \frac{\pi}{2}\right)(i=1$, $2, \cdots, n)$, if $\tan \theta_{1} \tan \theta_{2} \cdots \tan \theta_{n}=\sqrt{2^{n}}$, then $\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n}$ is not greater than $\lambda$. (2003 CMO Problem)
When $n=1$, $\cos \theta_{1}=\frac{1}{\sqrt{1+\tan ^{2} \theta_{1}}}=\frac{\sqrt{3}}{3}$, so $\lambda=\frac{\sqrt{3}}{3}$.

When $n=2$, it can be proven that
$$\cos \theta_{1}+\cos \theta_{2} \leqslant \frac{2 \sqrt{3}}{3}$$

and the equality holds when $\theta_{1}=\theta_{2}=-\arctan \sqrt{2}$. In fact, equation (1) $\Leftrightarrow \cos ^{2} \theta_{1}+\cos ^{2} \theta_{2}+2 \cos$ $\theta_{1} \cos \theta_{2} \leqslant \frac{4}{3}$, which means
$$\frac{1}{1+\tan ^{2} \theta_{1}}+\frac{1}{1+\tan ^{2} \theta_{2}}+2 \frac{1}{\sqrt{\left(1+\tan ^{2} \theta_{1}\right)\left(1+\tan ^{2} \theta_{1}\right)}} \leqslant \frac{4}{3}$$

From $\tan \theta_{1} \tan \theta_{2}=2$, we get
$$\text { equation (2) } \Leftrightarrow \frac{2+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}}{5+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}}+2 \frac{1}{\sqrt{5+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{1}}} \leqslant \frac{4}{3}$$

Let $x=\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}$, then
$$\text { equation (3) } \Leftrightarrow 2 \frac{1}{\sqrt{5+x}} \leqslant \frac{14+x}{3(5+x)}$$

which means
$$36(5+x) \leqslant 196+28 x+x^{2}$$

Clearly,
$$\text { equation (4) } \Leftrightarrow x^{2}-8 x+16=(x-4)^{2} \geqslant 0$$

Thus,
$$\lambda=\frac{2 \sqrt{3}}{3}$$

When $n \geqslant 3$, assume $\theta_{1} \geqslant \theta_{2} \geqslant \cdots \geqslant \theta_{n}$, then $\tan \theta_{1} \tan \theta_{2} \tan \theta_{3} \geqslant 2 \sqrt{2}$. Since $\cos \theta_{i}=\sqrt{1-\sin ^{2} \theta_{i}}7$, therefore,
$$\tan ^{2} \theta_{1} \geqslant \tan ^{2} \theta_{2}>\frac{7}{2}$$

So,
$$\cos \theta_{1} \leqslant \cos \theta_{2}0, \alpha \rightarrow 0$; then
$$\theta_{1}=\arctan -\frac{\sqrt{2^{n}}}{(\tan \alpha)^{n-1}}$$

Clearly, $\theta_{1} \rightarrow \frac{\pi}{2}$, thus
$$\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n} \rightarrow n-1$$

In summary, $\lambda=n-1$.