Odd Order Complete Graph is Eulerian

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

Then $K_n$ is Eulerian $n$ 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 an Eulerian trail, and so is traversable (although not Eulerian).

Historical Note
This was noted in 1809 by, who was unaware of Euler's more general result.

The remarkable point is that he gave an ingenious method for finding such a Eulerian circuit, which is a far from trivial exercise even for modestly sized complete graphs, e.g. those for $n < 100$.