if i understand the problem, you are looking for $d$ such that the $3$ cyclists $C_1,C_2,C_3$ can be closer to a distance less than $d$. for exemple $C_1C_2=d/3,C_2C_3=2d/3,C_1C_3=d$. is that what you mean?

Of course, they can be closer. We need to find such $d$ that it always possible.

But $\forall d>0$, the 3 cyclists can be closer with a distance less than $d$ so $d\to 0$

No, problem is to find such $d$ that for every starting positions and every speeds ( but different) will be moment for photoshoot.

let take $a,b$ respectively the relative speed of $C_1$ and $C_2$ compared to $C_3$ where $a>b$ $C_1$ speed $= C_3$ speed + $a$ and $C_2$ speed $=C_3$ speed + $b$ we take $t$ the elapsed time from the beginning of the course and $C_iC_j$ the clockwise distance between $C_i$ and $C_j$ Obviously $at=300A+r$ with $0\le r<300$ and $A$ integer $\iff$ $C_1C3=r$ now we take $n_1=a/300$ and $n_2=b/300$ Obviously $C_3C_i=300(n_it-[n_it])$ for $i=1,2$ where $[x]=$ the integer part of $x$ $x>y$ are integers when $t=\frac{x-y}{n_1-n_2}$ we have $C_3C_1=C_3C_2\iff$ $C_2$ and $C_3$ on the same position if this case $C_3C_1=C_3C_2=\frac{300(n_2x-n_1y)}{n_1-n_2}$ so $d:$ minimum distance $\le \frac{300(n_2x-n_1y)}{n_1-n_2}$ $\forall x,y$ integers satisfying $0\le \frac{n_2x-n_1y}{n_1-n_2}<1$

But what values in meters you get?

I claim that $\boxed{d=75}$. The hard part is guessing the ratio to be $0.25$ (this is motivated by Bezout's Lemma and/or the ease of construction). WLOG we can scale the track by $\tfrac{1}{300}$, work in the frame of reference of some runner, and linearly transform the time so that the second runner has position given by $p(t)=t-\lfloor t\rfloor$. If there were some construction with $d>0.25$, then there would exist a line with slope not $1$ that does not hit any of the green hexagons in the diagram. This can be shown false by considering when the line is first switches from being closer to the bottom of a hexagon than to the top of one or visa versa at integral time $t$. The construction to show that $d=0.25$ is tight follows by taking $p(t)=2t+0.5-\lfloor2t+0.5\rfloor$ $\blacksquare$

Attachments:

but if the speeds belonds to $\in \mathbb{Q}$, the $3$ cyclists can meet at one points, so $d=0$ we can even prove easly that $\forall$ speeds $\in \mathbb{R}$ the 3 cyclists can meet on a part on track $\le 1$ meter or even $1mm$

Right answer in $d=75$. Example: Let speeds of cyclists are $15,20,25$ m/sec. And let first and third cyclists are in point $A$ on track, and second in point $B$ , and $AB$=75. Then we can check, that in every moment distance between second and first or between second and third not less then $75m$.