Please do not answer this; this is a specific case of a problem from the Mathcamp Qualifying Quiz.

PM me or post here when the deadline is over because I don't want to lock the topic. It is a problem from another competition that is over. Till then, no solutions please.

Because the deadline is over for Mathcamp (because it is way past summer) I'll post my solution. It is possible for them to ride in this fashion. We can build a string of variables in the following fashion. Let $a_{1,1}=1$. Let $a_{n, n}=\text{lcm}(\{a_{n-1, k}: 1\leq k\leq n-1\})$ Let $a_{n, k}=a_{n, n}+a_{n-1, k}$ for $1\leq k\leq n-1$. In this way we build a triangular matrix of variables. It starts like this: $1$ $2, 1$ $4, 3, 2$ $16, 14, 13, 12$ $4384, 4382, 4381, 4380, 4368$ And so on, where the $k$th variable in the $n$th row is $a_{n, k}$. Note that $a_{n, k}>a_{n, k+m}$ if $m$ is positive. Now suppose there are $n$ horses. Let them all start at the point where they can meet, and let the $k$th horse have a speed of $a_{n, k}$ units per second. We claim that they will never meet, other than at the special point. We first prove that $a_{n, k}-a_{n, k+m}=\text{gcf}(a_{n, k}, a_{n, k+m})$. We do this by proving that both $a_{n, k}$ and $a_{n, k+m}$ are divisible by their difference; this certainly implies that the difference is the GCF. In fact, if we show that just one of them is divisible by their difference, then this also proves it. But this follows from the fact that $a_{n, k}=a_{n, n}+a_{n-1, n-1}+a_{n-2, n-2}+a_{n-3, n-3}+\ldots+a_{k, k}$ and $a_{n, k+m}=a_{n, n}+a_{n-1, n-1}+a_{n-2, n-2}+a_{n-3, n-3}+\ldots+a_{k+m, k+m}$ so $a_{n, k}-a_{n, k+m}=a_{k+m-1, k+m-1}+a_{k+m-2, k+m-2}+\ldots+a_{k, k}$ which doesn't depend on $n$. We can therefore replace $n$ with, say, $k+m$ and get $a_{n, k}-a_{n, k+m}=a_{k+m, k}-a_{k+m, k+m}=a_{k+m-1, k}$. This is the difference. Now we show that $a_{n, k}$ is divisible by this. We know that $a_{k+m, k+m}$ is the least common multiple of all the numbers $a_{k+m-1, c}$ and so it includes a factor of $a_{k+m-1, k}$ in it. Therefore so does $a_{k+m, k}$. And then so does $a_{k+m+1, k+m+1}$ and so does $a_{k+m+1, k}$. We can use induction to show that all $a_{c, c}$ and $a_{c, k}$ have $a_{k+m-1, k}$ as a factor if $c\geq k+m$. This includes $c=n$, so $a_{n, k}$ is divisible by $a_{m+k-1, k}=a_{n, k}-a_{n, m+k}$. Therefore this difference is the GCF of $a_{n, k}$ and $a_{n, m+k}$, which we wanted to prove. Now that that's out of the way, we can prove that no two horses ever are in the same place at the same time except at the special spot. Take horse number $k$ and $k+m$. They start at the same place to start, and every second horse $k$ speeds ahead $a_{n, k}$ units while horse $k+m$ only covers $a_{n, k+m}$ units. Thus every second, horse $k$ runs $a_{n, k}-a_{n, k+m}$ units more ahead of horse $k+m$ than it was already. The relative speed of horse $k$, from horse $k+m$'s perspective, is obviously $a_{n, k}-a_{n, k+m}$ units per second. Let's assume the track is $x$ units around. Then the first time horse $k$ and $k+m$ meet again after they start is after the distance between becomes $x$ units. This happens after $\frac{x}{a_{n, k}-a_{n, k+m}}$ seconds. After this time, horse $k$ has run $\frac{xa_{n, k}}{a_{n, k}-a_{n, k+m}}$ units. Now we've shown that $\frac{a_{n, k}}{a_{n, k}-a_{n, k+m}}$ is an integer, so horse $k$ has run an integer multiple of $x$ units, meaning it has run around the track an integer multiple of times and is at the special spot again. Therefore, any time two horses meet, they must be at the special spot and it is possible for them to go an arbitrary time like this. For your entertainment, here's a "video" of the case of 5 horses, with speeds 72, 78, 80, 81, 84 units per second. You can verify that the difference between any of these is equal to the GCF of them. To watch the "video", click the slider in the middle, then hold the right arrow key. The blue dot is the special point where the horses can cross. [geogebra]931f99a08ec962322d4732e6b35f69f37fa20ace[/geogebra]