Hunter 591 placed $n(n \geqslant 2)$ nets at different positions on a number line, the initial positions of the nets are denoted as $a_{1}, a_{2}$, $\cdots, a_{n}$. There is a bug between each pair of adjacent nets, and each net moves in the positive or negative direction at speeds $v_{1}, v_{2}, \cdots, v_{n}$. If two nets meet, the bug between them is captured, and the nets continue moving in their original manner, with a new bug flying between the two nets as they separate. This process continues indefinitely. What is the maximum number of bugs that can appear?
Let the maximum number of bugs caught be denoted as $f(n)$.
It is easy to see that $f(2)=2$.
It is evident that any two nets can meet at most once. Therefore, adding the $(n+1)$-th net can capture at most $n$ new bugs, meaning that during the movement, at most $n$ new bugs can be added, and one more bug can be added in the initial state, which gives us
$$
f(n+1) \leqslant f(n)+n+1 \text {. }
$$

Assume the movement of the $n$ nets that result in $f(n)$ bugs follows the conditions described in the problem. We will now prove that equality in (1) can be achieved.
Let $a_{n+1} = \max \left\{\left|a_{1}\right|,\left|a_{2}\right|, \cdots,\left|a_{n}\right|\right\} + 1$.
Suppose the time for the first meeting of the original $n$ nets is $t$, and let $v_{n+1} > \frac{\left|a_{n+1}\right|}{t}$, moving in the positive direction. It is not difficult to see that before the first meeting of the original $n$ nets, the $(n+1)$-th net has already met all the previous $n$ nets. This proves that
$$
f(n+1)=f(n)+n+1 \text {. }
$$

Calculating, we get $f(n)=\frac{n(n+1)}{2}-1$.