Let's define a tournament as follows:
"$T$ is a tournament if it is a directed graph where for each two vertices $x,y$, there exists exactly one edge connecting them" (Note that $x,y$ does not have to be distinct)
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Even though "$x$ beats $x$" does not make sense, let's just confine ourselves strictly to the definition.
Imagine a triangle with a loop on each vertex. Surely, this is a tournament by the definition. However, if we simply remove those 3 loops, then the triangle is still a tournament. And i think this is not actually a special example if a tournament is finite.
That is, let $T(V,E)$ be a finite tournament and $L$ be the set of all loops. If $L$ is nonempty, then is $|L|=|V|$?
If this is not true, does there exist a tournament $T$ such that $|L|\neq |V|$?