Example 45 Find the maximum positive integer $n$, such that there exist $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$ satisfying for any $1 \leqslant i<j \leqslant$ $n$, $\left(1+x_{i} x_{j}\right)^{2} \leqslant 0.99\left(1+x_{i}^{2}\right)\left(1+x_{j}^{2}\right)$.
Analyzing because (1) $\Leftrightarrow 100\left(1+x_{i} x_{j}\right)^{2} \leqslant 99\left(1+x_{i}^{2}\right)\left(1+x_{j}^{2}\right)$
$$
\begin{array}{l}
\Leftrightarrow\left(1+x_{i} x_{j}\right)^{2} \leqslant 99\left[\left(1+x_{i}^{2}\right)\left(1+x_{j}^{2}\right)-\left(1+x_{i} x_{j}\right)^{2}\right] \\
=99\left(x_{i}-x_{j}\right)^{2} \\
\Leftrightarrow\left(\frac{x_{i}-x_{j}}{1+x_{i} x_{j}}\right)^{2} \geqslant \frac{1}{99} .
\end{array}
$$

The structure on the left side reminds us of the trigonometric formula $\tan \left(\theta_{i}-\theta_{j}\right)=\frac{\tan \theta_{i}-\tan \theta_{j}}{1+\tan \theta_{i} \tan \theta_{j}}$.
Solution: Let the $n$ real numbers be $x_{i}=\tan \theta_{i}(i=1,2, \cdots, n), \theta_{i} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ and $\theta_{1} \leqslant \theta_{2} \leqslant \cdots \leqslant \theta_{n}$.
When $\theta_{n}-\theta_{1}>\frac{n-1}{n} \pi=\pi-\frac{\pi}{n}$, and because $\theta_{n}-\theta_{1}1-\left(\frac{\pi}{n}\right)^{2} \geqslant 1-\left(\frac{\pi}{32}\right)^{2}>1-0.1^{2}=0.99$. That is, when $n \geqslant 32$, for any $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$, there must exist two real numbers $x_{i}, x_{j}(1 \leqslant i0.99\left(1+x_{i}^{2}\right)\left(1+x_{j}^{2}\right)$.

Therefore, the maximum value of $n$ is $\leqslant 31$.
On the other hand, when $n=31$, take 31 real numbers as $x_{i}=\tan i \theta(i=1,2, \cdots, 31), \theta=\arctan \frac{1}{\sqrt{99}}$, then $\tan \theta=\frac{1}{\sqrt{99}}<\frac{10}{99}<\frac{\pi}{31}<\tan \frac{\pi}{31}$, hence $0<\theta<\frac{\pi}{31}$. For any $1 \leqslant i<j \leqslant 31$,
$$
\theta \leqslant \theta_{j}-\theta_{i} \leqslant 30 \theta<\frac{30}{31} \pi=\pi-\frac{\pi}{31}<\pi-\theta,
$$

Thus, $\tan ^{2}\left(\theta_{j}-\theta_{i}\right) \leqslant \tan ^{2} \theta=\frac{1}{99}$, i.e., $\left(\frac{x_{j}-x_{i}}{1+x_{i} x_{j}}\right)^{2} \leqslant \frac{1}{99}$,
From this, we can conclude that for any $1 \leqslant i<j \leqslant 31,\left(1+x_{i} x_{j}\right)^{2} \leqslant 0.99\left(1+x_{i}^{2}\right)\left(1+x_{j}^{2}\right)$.
In summary, the maximum value of $n$ is 31.