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.

Does every regular tournament have an antipodal pair of vertices? Furthermore, does every vertex have at least one antipode?

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.