Example 9 Given a positive integer $n$. Find the smallest positive number $\lambda$, such that for any $\theta_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2, \cdots, n)$, as long as
$$\tan \theta_{1} \tan \theta_{2} \cdots \tan \theta_{n}=2^{\frac{n}{2}}$$

then
$$\cos \theta_{1}+\cos \theta_{2}+\cdots+\cos \theta_{n} \leqslant \lambda$$
When $n=1$,
$$\cos \theta_{1}=\left(1+\tan ^{2} \theta_{1}\right)^{-\frac{1}{2}}=\frac{\sqrt{3}}{3}, \lambda=\frac{\sqrt{3}}{3}$$

Assume $n \geqslant 2$. Let $x_{i}=\tan ^{2} \theta_{i}\left(1 \leqslant_{i} \leqslant_{n}\right)$, then the given condition becomes
$$x_{1} x_{2} \cdots x_{n}=2^{n}$$

We need to find the smallest $\lambda$ such that
$$\sum \frac{1}{\sqrt{1+x_{i}}} \leqslant \lambda .$$

Without loss of generality, assume $x_{1}, x_{2}, \cdots, x_{n}$ where $x_{1}$ is the largest, then from equation (3): $x_{1} \geqslant 2$.

If $x_{2} \geqslant 2$, then it is exactly the same as the case $y \geqslant 2$ in Example 7 (where $x$ is $x_{1}$ and $y$ is $x_{2}$), we can conclude that $\sum \frac{1}{\sqrt{1+x_{i}}} = \phi(x_{2})$ is decreasing. Thus, we can decrease $x_{2}$ until $x_{2}=2$, at which point $\phi(x_{2})$ continues to increase. Similarly, we can assume $x_{3} \leqslant 2$, $x_{4} \leqslant 2, \cdots, x_{n} \leqslant 2$. Then, by the case $y \leqslant 2$ in Example 7, we can gradually adjust so that $x_{2}, x_{3}, \cdots, x_{n}$ are all adjusted to $\left(\frac{2^{n}}{x_{1}}\right)^{\frac{1}{n-1}}$. Therefore, we only need to find the smallest $\lambda$ such that
$$g(y)=\frac{1}{\sqrt{1+x}}+\frac{n-1}{\sqrt{1+y}} \leqslant \lambda$$

where $x \geqslant 2 \geqslant y, x y^{n-1}=2^{n}$.
$$\begin{array}{l}
g^{\prime}(y)=\frac{-(n-1)}{2(1+y)^{\frac{3}{2}}}+\frac{1}{2(1+x)^{\frac{3}{2}}} \frac{(n-1) 2^{n}}{y^{n}} \\
=\frac{n-1}{2 y}\left[\frac{x}{\sqrt{(1+x)^{3}}}-\frac{y}{\sqrt{(1+y)^{3}}}\right]
\end{array}$$
$g(y)$ is increasing $\Leftrightarrow g^{\prime}(y) \geqslant 0$
$$\begin{array}{l}
\Leftrightarrow \frac{x^{2}}{(1+x)^{3}} \geqslant \frac{y^{2}}{(1+y)^{3}} \\
\Leftrightarrow x^{2}(1+y)^{3}-y^{2}(1+x)^{3} \geqslant 0 . \\
\text { When } n=2 \text {, } x y=2^{2}, \\
y^{2}\left[x^{2}(1+y)^{3}-y^{2}(1+x)^{3}\right] \\
=4^{2}(1+y)^{3}-y(y+4)^{3} \\
=(2+y)(2-y)^{3} \geqslant 0 .
\end{array}$$

Therefore, $g(y)$ is increasing.
Thus, the maximum value is $g(2)=\frac{2}{\sqrt{3}}$, i.e., $\lambda=\frac{2}{\sqrt{3}}$.
When $n \geqslant 3$, $x y^{n-1}=2^{n}$,
$$\begin{array}{l}
y^{3 n-5}\left[x^{2}(1+y)^{3}-y^{2}(1+x)^{3}\right] \\
=2^{2 n} y^{n-3}(1+y)^{3}-\left(y^{n-1}+2^{n}\right)^{3} \\
=\left(2^{n}-y^{n}\right)\left(y^{2 n-3}+2^{n} y^{n-3}+3 \times 2^{n} y^{n-2}-2^{2 n}\right) \\
\leqslant\left(2^{n}-y^{n}\right)\left(2^{2 n-3}+2^{2 n-3}+3 \times 2^{2 n-2}-2^{2 n}\right) \\
=0
\end{array}$$

Therefore, $g(y)$ is decreasing.
When $y \rightarrow 0$, $g(y) \rightarrow n-1$, i.e., $\lambda=n-1$.