Two particles are moving in form of a unit circle.
One with a velocity v in the anticlockwise direction and one with velocity 3v in clockwise direction.
They meet at P and they again meet at R. Its very clear here that they meet (-1,0).
[Can be easily visualised by drawing the circle]
I shall generalise the problem which will make it harder which is the actual question.
Suppose we have $2n$ particles moving in a unit circle.
They are placed equidistant from each other having distance say $d$ from each other.
(Or You can say they make an equal angle of theta with each other consecutively)
Lets suppose when moving about the clockwise motion they have velocities $v,2v,3v,...,nv$.(in the first and the second quadrants).
Lets suppose when moving about the anticlockwise motion they have velocities $v,2v,3v,...,nv$.(in the fourth and the third quadrants).
Suppose at first they meet at point P. Find the coordinates of the point P.Let them meet again at R. What are the coordinates of the point R.
Suppose they meet for the nth time,what will be the coordinates of the point.
For simplicity, let's assume a pair of points start from $(1,0)$, one clockwise, and another counterclockwise.
For example, if red one goes at velocity $v$ counterclockwise, and blue one clockwise at velocity $3v$ or $2v$, they first meet at $90°$ or $120°$, respectively:
We can generalize this into one particle (red) going counterclockwise at velocity $pv$, and another (blue) clockwise at velocity $qv$.
The duration $t$ from the initial point to the first meeting point is such that when summed, the two particles have covered a full 360°, i.e. $$p\,v\,t + q\,v\,t = 360°$$ i.e. $$t = \frac{360°}{p\,v + q\,v}$$ This means that the first particle covers $$p\,v \frac{360°}{p\,v + q\,v} = 360° \frac{p\,v}{p\,v + q\,v} = 360° \frac{p}{p + q}$$ and the second $$q\,v \frac{360°}{p\,v + q\,v} = 360° \frac{q\,v}{p\,v + q\,v} = 360° \frac{q}{p + q}$$
In other words, the two particles will first meet at $$\varphi_1 = \begin{cases} \displaystyle 360° \frac{p}{p + q}, & p \le q \\ \displaystyle -360° \frac{q}{p + q}, & q \lt p \end{cases}$$ and since the situation repeats -- that is, we can treat the first meeting as a new initial positions --, the following meeting points will simply occur at multiples of that angle, $$\varphi_n = n \varphi_1 , \qquad 2 \le n \in \mathbb{N}$$ Or, because the starting point $\varphi_0 = 0°$, we can simplify it as $$\varphi_i = \begin{cases} \displaystyle 360° \frac{i\;p}{p + q}, & p \le q \\ \displaystyle -360° \frac{i\;q}{p + q}, & p \gt q \end{cases}\tag{1a}\label{NA1a}$$ or in Cartesian coordinates, $$\begin{cases} \displaystyle x_i = \cos\left(\frac{2 \pi \, i \, p}{p + q}\right) \\ y_i = \begin{cases} \displaystyle \sin\left(\frac{2 \pi \, i \, q}{p + q}\right), & p \le q \\ \displaystyle -\sin\left(\frac{2 \pi \, i \, q}{p + q}\right), & p \gt q \end{cases} \end{cases}\tag{1b}\label{NA1b}$$
In OP's case, we have $2n$ particles, with $p_i = i$, $q_i = i$, $i = 1..n$. As angles a function of time, we have $$\begin{aligned} \varphi_{i}(t) &= \alpha i t \qquad & \text{for p particles} \\ \theta_{i}(t) &= -\alpha i t \qquad & \text{for q particles} \end{aligned}$$ where $\alpha$ is some angular velocity constant, that depends on $v$ and the choice of time units.
Essentially, we are looking at the other half of the full period of the system, compared to the question OP posed.
Note that after the slowest particles have done a full turn, the next slowest particles have done two full turns, and the fastest particles have done $n$ full turns; but all particles are back at their initial positions.
Numerically, I found out that because there is no particle with zero speed, starting at the same point with angular velocities $\pm \alpha$, $\pm 2 \alpha$, ..., $\pm n\,\alpha$, the particles are never uniformly distributed along the circle: $$\begin{array}{l|ccc|c} n & \varphi_1 & \delta_{max} & \delta_{min} & \delta \\ \hline 2 & 154.28° & 102.85° & 51.42° & 51.43° \\ 3 & 108.00° & 72.00° & 36.00° & 36.00° \\ 4 & 83.08° & 55.38° & 27.69° & 27.69° \\ 5 & 67.50° & 45.00° & 22.50° & 22.50° \\ 6 & 56.84° & 37.89° & 18.95° & 18.95° \\ 7 & 49.09° & 32.73° & 16.36° & 16.36° \\ 8 & 43.20° & 28.80° & 14.40° & 14.40° \\ 9 & 38.57° & 25.72° & 12.86° & 12.86° \\ 10 & 34.8° & 23.2° & 11.6° & 11.6° \\ 20 & 17.7° & 11.8° & 5.9° & 5.9° \\ \end{array}$$ where $\varphi_1$ denotes the angle of the slowest particle, $\delta_{max}$ the largest angular separation between neighboring particles (without any other particles in between), and $\delta_{min}$ the smallest angular separation between neighboring particles, at minimum possible $\delta = \delta_{max} - \delta_{min}$.
(The numerical test considered angles modulo 360°, and ignored the "velocity order" of the particles, i.e. considered only particle pairs angular differences, where there were no other particles in between, and took special care to handle the pair that crossed 360° correctly. I did this, because I thought there might be a case where the particles were uniformly distributed, but their angular positions in different order than their velocities.)
Note that $n=1$ is an exception, since there is just one pair of particles, having the same velocity. When they have travelled one quarter of the unit circle, they are opposite each other, 90° separated from the starting point. After they travel another quarter circle, they are at the same point, 180° from their starting point. This repeats ad infinitum.
Thus, it looks like the answer to OP's question for $2 \le n \in \mathbb{N}$ is never. If the particles start uniformly distributed along the unit circle, with angular velocities $\pm \alpha$, $\pm 2 \alpha$, ..., $\pm n \alpha$, they will never meet in a single point.