Has the following problem, resembling the lonely runner conjecture, been studied? Given $n$, what is the smallest value $\delta_n$ satisfying the following:

For any group of $n$ runners with constant but distinct speeds,
starting from the same point and running clockwise along the unit
circle,  there exists a moment, at which the length of each of $n$
empty circle segments between two consecutive runners does not exceed
$\delta_n$.

More formally, for $n \in \mathbb{N}$, $M=\{m_1,\dots,m_n\} \subset \mathbb{N}$, and $t \in [0,1]$, we sort the fractional parts of the values $tm_i$ in ascending order: $0 \le \{tm_{i_1}\} \le\dots\le \{tm_{i_n}\} \le 1$. Then we consider the longest empty segment between two consecutive 'runners':
$$ \text{LongestArc}(M,t) := \max\Big\{ \{tm_{i_2}\}-\{tm_{i_1}\}, \dots, \{tm_{i_n}\}-\{tm_{i_{n-1}}\}, 1+\{tm_{i_1}\}-\{tm_{i_n}\} \Big\}.$$
Finally, we put
$$ \delta_n := \mathop{\smash{\mathrm{sup}}}_{|M|=n} \inf_{t \in [0,1]} \text{LongestArc}(M,t).$$
Has the problem of finding these values (or, perhaps, an equivalent one) been studied earlier? I have looked through some papers related to the Lonely Runner Conjecture (including this one) but found nothing.
It is clear that $\delta_2=\frac12$. With a little more effort goes $\delta_3=\frac25$, one of the 'extremal' sets here is $M=\{0,1,3\}$. One may check that $\delta_4 \ge \frac13$ by taking $M=\{0,1,2,4\}$. I suspect this bound to be tight. Maybe someone sees a counterexample or a short argument, why is that so?
For large $n$, I believe I can show that $\delta_n \ge \frac{\log n -O(\log \log n)}{n}$. However, I don't have any (decent) upper bound on $\delta_n$ apart from the fact that $\delta_n=o(1)$ as $n \to \infty$.
Finally, I found that the 'dual' problem concerning the values
$$ \epsilon_n := \inf_{|M|=n} \mathop{\smash{\mathrm{sup}}}_{t \in [0,1]} \text{LongestArc}(M,t)$$
has earlier been considered on MathOverflow (not in the literature, though). One may also state two more 'dual' extremal problems related to the shortest arc (instead of the longest one), but I haven't found anything published about them as well.
I would appreciate any related reference!
 A: I think the answer to the question as posed is "yes", see Konyagin-Ruzsa-Schlag (2011)
https://gauss.math.yale.edu/~ws442/papers/DILATES.pdf
It looks like nothing beyond the bounds discovered in the comments and Fedja's answer above is known.
A: OK, since Will thinks it is worth posting, here it goes. This is not an answer, just setting up the triviality level. As usual, the circle will be identified with $\mathbb R/\mathbb Z$.
Let $\psi$ be the piecewise linear $1$ periodic function that is $0$ at $0$, raises to $1$ at $\delta/2$, drops down to $0$ at $\delta$ and then stays $0$ throughout the rest of the period (the well-known "triangle hump", just shifted forward from its usual center). We have the Fourier expansion
$$
\psi(x)=\frac{\delta}2+\sum_{k\ne 0}c_ke^{2\pi ikx}
$$
with $\sum_{k\ne 0}|c_k|=1-\frac\delta 2<1$.
We want to find $t$ such that for every runner moving at the speed $v_0\in V$ ($V$ is the set of all speeds), we have
$$
\Psi_{v_0}(t)=\sum_{v\in V} \psi(vt-v_0t)\ne 0\,.
$$
This will mean that there is at least one runner in the interval of length $\delta$ ahead of every runner, so the maximal gap is at most $\delta$.
However
$$
\Psi_{v_0}(t)=\frac{\delta n}2+\sum_{k\ne 0}c_ke^{-2\pi i kv_0t}F(kt)
\\
\ge 
\frac{\delta n}2-\sum_{k\ne 0}|c_k||F(kt)|\ge
\frac{\delta n}2-\sqrt{\sum_{k\ne 0}|c_k||F(kt)|^2}
$$
where $F(t)=\sum_{v\in V}e^{2\pi i vt}$.
Note now that for each $k\ne 0$, the mean value of $|F(kt)|^2$ (in the sense of almost-periodic functions, i.e., $\lim_{T\to\infty}\int_{-T}^T\dots\,dt$) is $n$, so the mean value of the sum under the square root is $<n$. Thus, we are fine if $\frac{\delta n}2-\sqrt n\ge 0$, i.e., if $\delta\ge \frac 2{\sqrt n}$.
Edit: I guess that in the view of mathworker21's comments, I should emphasize a bit where exactly the real difficulty is. To this end I'll recast the proof into another language avoiding the Fourier transform altogether though it is actually the same proof.
Let $\delta>0$. Consider the functions $\varphi_v(x,t)$ that are 1-periodic in $x$ and given by $\chi_{[0,\delta]}(vt-x)-\delta$ on the period and $\Phi(x,t)=\sum_v\varphi_v(x,t)$. If for some $t$ we have a $2\delta$-gap, then $$
\int_0^1 |\Phi(x,t)|^2\,dx\ge \delta(-n\delta)^2=n^2\delta^3\,.
$$
Thus, if we have that gap for all $t$, we must have
$$
\operatorname{Mean}_t\int_0^1 |\Phi(x,t)|^2\,dx\ge n^2\delta^3
$$
but we have the joint in $x,t$ ortogonality of $\varphi_v$, so the left hand side is just $n\delta(1-\delta)$.
The issue is that when $\delta<n^{-1/2}$, the $n\delta$ bump doesn't rise in the $L^2$ integral sense above standard "pure noise level" $\sqrt{n\delta}$ on $[0,1]\ni x$. Trying to use higher $L^p$ norms is problematic for two reasons. First, the integrals of long products now depend on linear dependences between $v$'s with integer coefficients and second, if all speeds are, say, integers from $1$ to $n$, the initial overcrowding interval of length $\frac \delta n$ will already contributes $n^{p-1}\delta$ to that norm, which is way too much. We really need to get rid of (or, at least, suppress) the positive part of $\Phi$ before averaging and that is not easy with polynomial functions, so one needs some new trick or even some new idea to get something non-trivial (which is more or less the definition of "non-trivial").
