Solution:
The order of the ants along the line does not change; denote by $v_{1}, v_{2}, \ldots, v_{n}$ the respective speeds of ants $1,2, \ldots, n$ in this order. If $v_{i-1}<v_{i}>v_{i+1}$ for some $i \in\{2, \ldots, n-1\}$, then, at each stage, ant $i$ can catch up with ants $i-1$ or $i+1$ irrespective of the latters' directions of motion, so the number of collisions is infinite. Hence, if the number of collisions is finite, then, up to switching the direction definining the order of the ants, (i) $v_{1} \geqslant \cdots \geqslant v_{n}$ or (ii) $v_{1} \geqslant \cdots \geqslant v_{k-1}>v_{k} \leqslant \cdots \leqslant v_{n}$ for some $k \in\{2, \ldots, n-1\}$. We need the following observation:
Claim. If $v_{1} \geqslant \cdots \geqslant v_{m}$, then ants $m-1$ and $m$ collide at most $m-1$ times.

Proof. The proof goes by induction on $m$, the case $m=1$ being trivial. Since $v_{m-1} \geqslant v_{m}$, ants $m-1$ and $m$ can only collide if the former is moving towards the latter. Hence, between successive collisions with ant $m$, ant $m-1$ must reverse direction by colliding with ant $m-2$. Since ants $m-1$ and $m-2$ collide at most $(m-1)-1=m-2$ times by the inductive hypothesis, ants $m$ and $m-1$ collide at most $(m-2)+1=m-1$ times.

Hence, in case (i), there are at most $0+1+\cdots+(n-1)=n(n-1) / 2$ collisions. In case (ii), applying the claim to ants $1,2, \ldots, k$ and also to ants $n, n-1, \ldots, k$ by switching their order, the number of collisions is at most $k(k-1) / 2+(n-k+1)(n-k) / 2=n(n-1) / 2-(k-1)(n-k)<n(n-1) / 2$.

Now take a coordinate $x$ along the line, and put ants at $x=1,2, \ldots, n$ with positive initial velocities and speeds $v_{1}=\cdots=v_{n-1}=1, v_{n}=\varepsilon$, for some $\varepsilon$. For $\varepsilon=0$, collisions occur according to the pattern shown below for $n=5$, which clearly extends to all values of $n$ in such a way that ants $m$ and $m+1$ collide exactly $m$ times for $m=1,2, \ldots, n-1$. This yield $1+2+\cdots+(n-1)=n(n-1) / 2$ collisions in total. For all sufficiently small $\varepsilon>0$, the number of collisions remains equal to $n(n-1) / 2$.

This shows that the upper bound obtained above can be attained. If the number of collisions is finite, the largest possible number of collisions is therefore indeed $n(n-1) / 2$.

Solution:
We show that there are at most $n(n-1) / 2$ collisions if the number of collisions is finite as in Solution 1.

To show that the upper bound of $n(n-1) / 2$ collisions can be attained, we construct, inductively, an example of $n$ ants colliding $n(n-1) / 2$ times, the speeds of the ants decrease from left to right, and after all collisions all ants move towards the left, with the possible exception of the rightmost ant. In every case, we will label the ants $1,2, \ldots, n$ from left to right. For $n=1$ this is trivial. For $n \geqslant 2$, we use the construction for $n-1$ ants (now labelled $2,3, \ldots, n$ ). We add ant 1 on the left, moving towards the right, faster than all other ants (so that the speeds of the ants still decrease from left to right), and in such a way that its first collision (with ant 2) happens after all $(n-1)(n-2) / 2$ collisions of the other $n-1$ ants. Now the following events happen (in this order) for $i=1,2, \ldots, n-2$ : ants $i$ and $i+1$ collide, after which ant $i$ moves to the left and ant $i+1$ moves to the right. These collisions do happen because the speeds of the ants decrease from left to right. Then ants $n-1$ and $n$ also collide, resulting in ant $n-1$ moving to the left. This shows that there are (at least) $(n-1)(n-2) / 2+(n-1)=n(n-1) / 2$ collisions. There are in fact no more collisions since the speeds of the ants decrease from left to right; alternatively, this follows from the upper bound proved previously. Since all ants except ant $n$ are moving towards the left after the collisions, this completes the inductive construction.