14. (2003 China Mathematical Olympiad) Given a positive integer $n$, find the smallest positive number $\lambda$, such that for any $\theta_{i} \in\left(0, \frac{\pi}{2}\right)(i=1$, $2, \cdots, n)$, if $\tan \theta_{1} \cdot \tan \theta_{2} \cdots \tan \theta_{n}=2^{\frac{n}{2}}$, then $\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n}$ is not greater than $\lambda$.
14. When $n=1$, $\cos \theta_{1}=\left(1+\tan ^{2} \theta_{1}\right) \frac{1}{2}=\frac{\sqrt{3}}{3}$, we have $\lambda=\frac{\sqrt{3}}{3}$. 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,
$$
\text { Equation (1) } \Leftrightarrow \cos ^{2} \theta_{1}+\cos ^{2} \theta_{2}+2 \cos \theta_{1} \cdot \cos \theta_{2} \leqslant \frac{4}{3} \text {, }
$$

which means $\frac{1}{1+\tan ^{2} \theta_{1}}+\frac{1}{1+\tan ^{2} \theta_{2}}+2 \sqrt{\frac{1}{\left(1+\tan ^{2} \theta_{1}\right)\left(1+\tan ^{2} \theta_{2}\right)}} \leqslant \frac{4}{3}$. 
By $\tan \theta_{1} \cdot \tan \theta_{2}=2$ we get
$$
\text { Equation (2) } \Leftrightarrow \frac{2+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}}{5+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}}+2 \sqrt{\frac{1}{5+\tan ^{2} \theta_{1}+\tan ^{2} \theta_{2}}} \leqslant-\frac{4}{3} \text {. }
$$

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

which means $36(5+x) \leqslant 196+28 x+x^{2}$.
Clearly, Equation (4) $\Leftrightarrow x^{2} \quad 8 x+16=(x-4)^{2} \geqslant 0$.
Thus, $\lambda=\frac{2 \sqrt{3}}{3}$.
When $n \geqslant 3$, assume $\theta_{1} \geqslant \theta_{2} \geqslant \cdots \geqslant \theta_{n}$, then
$$
\tan \theta_{1} \cdot \tan \theta_{2} \cdot \tan \theta_{3} \geqslant 2 \sqrt{2} \text {. }
$$

Since $\cos \theta_{i}=\sqrt{1-\sin ^{2} \theta_{i}}0, \alpha \rightarrow 0$,
$\theta_{1}=\arctan \frac{2^{\frac{n}{2}}}{(\tan \alpha)^{n-1}}$, clearly $\theta_{1} \rightarrow \frac{\pi}{2}$, thus
$$
\cos \theta_{1}+\cos \theta_{2}+\cos \theta_{3}+\cdots+\cos \theta_{n} \rightarrow n-1 \text {. }
$$

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