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

This leads us to think of the tangent formula for the difference of two angles: Let $x_{i}=\tan \theta_{i}, \tan \theta=\frac{1}{\sqrt{99}}$, then the above inequality is equivalent to
$$
\tan ^{2}\left(\theta_{j}-\theta_{i}\right) \geqslant \tan ^{2} \theta \text {. }
$$

On one hand, for any $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$, let $x_{i}=\tan \theta_{i}\left(-\frac{\pi}{2}\frac{n-1}{n} \pi=\pi-\frac{\pi}{n}\right)$, when $\theta_{n}-\theta_{1}1-\left(\frac{1}{10}\right)^{2}=0.99$, i.e., when $n \geqslant 32$, for any $n$ real numbers $x_{1}, x_{2}, \cdots, x_{n}$, there must exist two real numbers $x_{i}, x_{j}$ such that
$$
\left(1+x_{i} x_{j}\right)^{2}>0.99\left(1+x_{i}^{2}\right)\left(1+x_{j}^{2}\right) \text {. }
$$

Therefore, when $n \geqslant 32$, there do not exist $n$ real numbers that satisfy the given condition.
On the other hand, take 31 real numbers $x_{i}=\tan (i \theta)(i=1,2, \cdots, 31)$, where $\theta=\arctan \frac{1}{\sqrt{99}}$, then $\tan \theta=$ $\frac{\sqrt{99}}{99}<\frac{10}{99}<\frac{\pi}{31}<\tan \frac{\pi}{31}$, i.e., $0<\theta<\frac{\pi}{31}$. So, when $1 \leqslant i<j \leqslant 31$,
$$
\theta \leqslant(j-i) \theta \leqslant 30 \theta<\frac{30 \pi}{31}=\pi-\frac{\pi}{31}<\pi-\theta .
$$

Thus, $\tan ^{2}(j-i) \theta \geqslant \tan ^{2} \theta=\frac{1}{99}$,
i.e., $\left(\frac{x_{j}-x_{i}}{1+x_{i} x_{j}}\right)^{2} \geqslant \frac{1}{99}$.
Therefore, $\left(1+x_{i} x_{j}\right)^{2} \leqslant 0.99\left(1+x_{i}^{2}\right)\left(1+x_{j}^{2}\right)$.
Hence, there exist 31 real numbers $x_{1}, x_{2}, \cdots, x_{31}$ that satisfy the given condition.
In summary, the maximum value of $n$ is 31.