Odd Order Complete Graph is Eulerian

Theorem
Let $$K_n$$ be the complete graph of $$n$$ vertices.

Then $$K_n$$ is Eulerian iff $$n$$ is odd.

If $$n$$ is even, then $$K_n$$ is traversable iff $$n = 2$$.

Proof
From the definition, the complete graph $$K_n$$ is $n-1$-regular.

That is, every vertex of $$K_n$$ is of degree $$n-1$$.

Suppose $$n$$ is odd. Then $$n-1$$ is even, and so $$K_n$$ is Eulerian.

Suppose $$n$$ is even. Then $$n-1$$ is odd.

Hence for $$n \ge 4$$, $$K_n$$ has more than $$2$$ odd vertices and so can not be traversable, let alone Eulerian.

If $$n = 2$$, then $$K_n$$ consists solely of two odd vertices (of degree $$1$$).

Hence, by Condition for Graph to be Traversable (or trivially, by inspection), $$K_2$$ has a Eulerian path, and so is traversable (although not Eulerian).