Prove that when $x^{2} \geqslant a_{1}\left(a_{1}-1\right)$, due to $\sum \frac{1}{a_{i}} \leqslant 1$, we have
$$\begin{aligned} \left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} & \leqslant\left(\sum_{i=1}^{n} \frac{1}{2 a_{i}|x|}\right)^{2}=\frac{1}{4 x^{2}}\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right)^{2} \\ & \leqslant \frac{1}{4 x^{2}} \leqslant \frac{1}{2} \cdot \frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}} \end{aligned}$$

When $x^{2}<a_{1}\left(a_{1}-1\right)$, by the Cauchy-Schwarz inequality, we have
$$\begin{aligned} \left(\sum_{i=1}^{n} \frac{1}{a_{i}^{2}+x^{2}}\right)^{2} & \leqslant\left(\sum_{i=1}^{n} \frac{1}{a_{i}}\right) \sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \\ & \leqslant \sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \end{aligned}$$

For positive integers $a_{1}<a_{2}<\cdots<a_{n}$, we have $a_{i+1} \geqslant a_{i}+1, i=1,2, \cdots, n-1$, and
$$\begin{aligned} \frac{2 a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} & \leqslant \frac{2 a_{i}}{\left(a_{i}^{2}+x^{2}+\frac{1}{4}\right)^{2}-a_{i}^{2}} \\ & =\frac{1}{\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}}-\frac{1}{\left(a_{i}+\frac{1}{2}\right)^{2}+x^{2}} \\ & \leqslant \frac{1}{\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}}-\frac{1}{\left(a_{i+1}-\frac{1}{2}\right)^{2}+x^{2}}, i=1,2, \cdots, n-1 \end{aligned}$$

Similarly,
$$\begin{aligned} \frac{2 a_{n}}{\left(a_{n}^{2}+x^{2}\right)^{2}} & \leqslant \frac{1}{\left(a_{n}-\frac{1}{2}\right)^{2}+x^{2}}-\frac{1}{\left(a_{n}+\frac{1}{2}\right)^{2}+x^{2}} \\ & \leqslant \frac{1}{\left(a_{n}-\frac{1}{2}\right)^{2}+x^{2}} \end{aligned}$$

Therefore,
$$\begin{aligned} \sum_{i=1}^{n} \frac{a_{i}}{\left(a_{i}^{2}+x^{2}\right)^{2}} \leqslant & \frac{1}{2} \sum_{i=1}^{n-1}\left[\frac{1}{\left(a_{i}-\frac{1}{2}\right)^{2}+x^{2}}-\frac{1}{\left(a_{i+1}-\frac{1}{2}\right)^{2}+x^{2}}\right] \\ & +\frac{1}{\left(a_{n}-\frac{1}{2}\right)^{2}+x^{2}} \\ \leqslant & \frac{1}{2} \cdot \frac{1}{\left(a_{1}-\frac{1}{2}\right)^{2}+x^{2}} \leqslant \frac{1}{2} \cdot \frac{1}{a_{1}\left(a_{1}-1\right)+x^{2}} \end{aligned}$$

Thus, the proposition is proved.