The answer is $(n-1)m'$ where $m'$ is the smallest number such that $\frac{m'(n-2)}{2n}$ is an integer.
More precisely, the answer is
* $2n(n-1)$ when $n$ is odd
* $n(n-1)$ when $n$ is divisible by 4
* $\frac{n(n-1)}{2}$ when $n-2$ is divisible by 4. (but the $n/2$-th point will be the one moving in reverse, not $p_0$)

We first introduce some setup for convenience. We treat points on the circle as $[0, 2n)$ and we use $r = r + 2n$ for any real to refer to points on the circle for convenience. For example, we can use point $-1$ to mean $2n - 1$ etc. Now, we also assume that it takes any particle $2n$ units of time to go around the entire circle. Thus, if a particle at $0$ moves clockwise for time $t$ then it would reach point $t$.

Initially, we let the particles be $p_0, \dots, p_{n-1}$ with $p_i$ at point $2i$.

Now, moving clockwise means the value is increasing and moving anti-clockwise means value is decreasing and finally we assume $p_0$ is the point initially moving anti-clockwise.

Now, let $t_1 > 0$ be the total time when for the first time, all particles return to their initial positions.
Correspondingly let $t_2 > 0$ be the first time when all particles are equally spaced apart i.e. 2 units apart.

Now, clearly $\frac{t_1}{t_2}$ is an integer. Thus, we try to find $t_2$.

Observe that if we replace each collision event with the two particles passing through each other, $t_2$ does not change. So for the purposes of calculating $t_2$, we can assume they indeed pass through. Thus, at time $t_2$, there are particles at $-t_2, 2+t_2, 4+t_2, \dots, 2n-2+t_2$. But for them to be equally spaced, this sequence must be $t_2, 2+t_2, 4+t_2, \dots, 2n-2+t_2$. Thus, modulo $2n$, $t_2 = -t_2$, and the minimum value of $t_2$ that makes this possible is $t_2 = n$. Now, we can set $t_1 = n \cdot t'_1$.

Alternatively, one could have observed that we can look at the positions relative to the clockwise moving particles. Then we just have the anti-clockwise particle moving at speed 2 so it must need $n$ units of time to return to original position and make $n-1$ collisions in this period.

Now, let us analyze what happens at time $n$. There are now particles in positions $n, n+2, \dots, n-2$. Observe that cyclic order of particles must be preserved as collisions never alter it and finally the collisions only happen in $(1, n)$ after the first collision of $p_0$ at point $2n - 1$ with $p_n$. Thus, $p_0$ does not have any more collisions in the next $n-1$ units of time. Thus, $p_0$ must be at position $2n - 1 + (n-1) = n - 2$. Now, since cyclic order is preserved, $p_i$ would be at position $2i + n - 2$. Thus, everything has cyclically moved forward by $n-2$.

Thus, in time $n$, every particle moves forward by $n-2$ units and there are $n-1$ collisions. Thus, in time $nk$, we have $(n-1)k$ collisions and every particles moves forward by $(n-2)k$ units. Thus, if every particles to their initial position iff $2n \mid (n-2)k$. This is exactly what we desired! $\square$

Assume the circumference is 2 units, and the particles have speed 1 unit per second. Call a position "good" if the particles are equally spaced. We consider the distance of the particles as positive in the anti-clockwise direction and negative in the clockwise direction.

We first claim that if the position is good, the distance travelled by each particle is the same. Indeed, as the relative positions of the particles does not change (and the particles don't pass through each other), the distance travelled by any particle is sandwiched between the distances travelled by its neighbors, so by cyclicity all the distances must be equal.

Thus, the original position is repeated iff the final position is good and the sum of distances travelled by the particles is $2nl$ for some integer $l$. Since we only care about sum of distances and not the individual particles anymore, we consider the particles to be indistinguishable, and assume that they pass through each other when they collide.

As in the first solution, we can show that the first good position happens after $t_2 = 1$ second, with number of collisions being $n-1$ so far. In that 1 unit of time, one particle has travelled a clockwise distance of 1, with $n-1$ others have travelled an anti-clockwise distance of 1. This implies that the total distance travelled in time $t_2$ is $n-2$.

Thus, if $\frac{t_1}{t_2} = m$, then number of collisions is $m(n-1)$, and total distance travelled is $m(n-2)$. Thus the particles return to the original position iff $2n \mid m(n-2)$, which gives the required answer as seen in the first solution. $\square$