Example 11 Find the largest 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
$$\left(1+x_{i} x_{j}\right)^{2} \leqslant 0.9\left(1+x_{i}^{2}\right)\left(1+x_{j}^{2}\right)$$
Solve
$$\left(1+x_{i} x_{j}\right)^{2} \leqslant 0.9\left(1+x_{i}^{2}\right)\left(1+x_{j}^{2}\right)$$

This is equivalent to
$$0.1\left(x_{i} x_{j}+1\right)^{2} \leqslant 0.9\left(x_{i}-x_{j}\right)^{2}$$

Which is also
$$\left|x_{i} x_{j}+1\right| \leqslant 3\left|x_{i}-x_{j}\right|$$

Let $x_{i}=\tan \alpha_{i}(1 \leqslant i \leqslant n)$, without loss of generality, assume $0 \leqslant \alpha_{1}\pi$. So
$$9 \theta<\pi<10 \theta$$

Since $\pi-\theta \geqslant \alpha_{n}-\alpha_{1} \geqslant(n-1) \theta$, then $n \leqslant 9$.
When $\alpha_{1}=0, \alpha_{2}=\theta, \alpha_{3}=2 \theta, \cdots, \alpha_{9}=8 \theta$, the equality can be achieved. Therefore, the maximum value of $n$ is 9.