Odd Order Complete Graph is Eulerian
If $n$ is even, then $K_n$ is traversable iff $n = 2$.
Suppose $n$ is odd. Then $n-1$ is even, and so $K_n$ is Eulerian.
Suppose $n$ is even. Then $n-1$ is odd.
If $n = 2$, then $K_n$ consists solely of two odd vertices (of degree $1$).
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, for example those for $n < 100$.