Does there exist a tournament with loops where the numbers of loops is strictly smaller than the number of vertices? 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 $|L|\neq |V|$?

There's a reason you earned the tumbleweed badge for this question.  It's essentially asking:

"If I define a tournament as [something] is it possible to have a
  tournament satisfying [some conditions]?"

It would be better to decide whether or not you want [some condition] to be true, then define a tournament based on that.
But, to answer the question, the definition says "for each two vertices $x,y$, there exists exactly one edge connecting them", which, since you've specified that $x$ may equal $y$, implies there's a loop on every vertex to cover all the $x=y$ cases.
