Let $T$ be a regular tournament, and $u \in V(T)$. Let $Out(u) \subset V(T)$ denote the set of vertices such that the edges between $u$ and them go out of $u$. Similarly define $In(u)$. Let two distinct vertices $u,v$ be called $\textbf{antipodal}$ if $Out(u) \setminus v = In(v) \setminus u$ (and thus also $In(u) \setminus v = Out(v) \setminus u$). Put more concretely, if $w$ is any other vertex, then either $u$ goes into $w$ AND $w$ goes into $v$, or vice versa.


Given $u$, let $v$ be called an $\textbf{antipode}$ of $u$ if $u$ and $v$ are antipodal. Note that this is poor notation as $u$ could have more than one antipode.



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> Does every regular tournament have an antipodal pair of vertices? Furthermore, does every vertex have at least one antipode?
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Note that the first problem is equivalent to showing that a regular tournament on $2n+1$ vertices contains an induced subgraph on $2n-1$ vertices that is a regular tournament.


The motivation for this problem is the following observation (if it's correct). $T$ has a size $2n-1$ subgraph that is a regular tournament if and only if, for any partition of $2n+1$ into odd numbers, $T$ can be partitioned into regular tournaments as subgraphs of sizes corresponding to the partition.


Perhaps this result is too strong to be true, I have only verified it for some very small examples.



About regular induced subgraphs of regular tournaments
There are regular tournaments [here](https://users.cecs.anu.edu.au/%7Ebdm/data/digraphs.html) and if you choose one at random except on the smallest sizes you will find a counterexample with high probability.


Enumeration gives a bound (which could be turned into an accurate estimate but I won't). Let $T(n)$ be the number of (labelled) regular tournaments with $n$ vertices. To make a regular tournament with $n+2$ vertices and an antipodal pair, take a regular tournament with $n$ vertices and add to it to new vertices $v\to w$. There are less than $2^n$ ways to add the edges between the new vertices and the old vertices such that $v,w$ are antipodal, then you can multiply by $n^2$ for the labels of the new vertices. So the number of regular tournaments on $n+2$ vertices that have an antipodal pair is at most $n^2 2^n T\_n$. However, $T\_{n+2}$ is approximately $2^{2n}n^{-1}T\_n$. So only an exponentially small fraction of regular tournaments have an antipodal pair.


Ref for enumeration: B. D. McKay, [The asymptotic numbers of regular tournaments, eulerian digraphs and eulerian oriented graphs](https://users.cecs.anu.edu.au/%7Ebdm/papers/rt.pdf), Combinatorica, 10 (1990) 367-377.


