When can a graph be drawn without lifting the pen?

1. Among all trails, take a longest one, $T=(v_0,v_1,\dots,v_m)$.

*$T$ is closed.* Suppose $v_m\ne v_0$. Each time $T$ passes through $v_m$ (other than the end) it uses two of its edges, and the last arrival uses one more, so $T$ uses an odd number of edges at $v_m$. Since $\deg v_m$ is even, there is an unused edge at $v_m$, and $T$ can be extended: contradiction. So $v_m=v_0$.

*$T$ uses all edges.* Suppose not. Because the graph is connected, some unused edge $e$ has an endpoint $w$ on $T$: take an unused edge $f$ and a path from a vertex of $T$ to an endpoint of $f$ ending with $f$; the first unused edge $e$ on this path starts at a vertex reached by used edges only. Rotate the closed trail $T$ so that it starts and ends at $w$, then append $e$: this is a longer trail, a contradiction.

2. Let $a,b$ be the odd vertices. Add a new edge $ab$ (a parallel edge if needed; the proof of part 1 works verbatim for multigraphs). All degrees are now even, so by part 1 there is a closed trail through all edges. Removing the new edge from it gives a trail from $a$ to $b$ through all the original edges exactly once.

3. All degrees of $K_n$ equal $n-1$. If $n$ is odd, the degrees are even and $K_n$ is connected, so a closed trail exists. If $n$ is even and $n\ge2$, every vertex has odd degree and a closed trail using all edges cannot exist. So the answer is: exactly the odd $n$ (for $n=1$ there are no edges and the empty trail works).

Remark. The converse of 1 is easy, so the condition is an equivalence. This is Euler's theorem (Königsberg bridges, 1736).

Lesson. Extremal choice: take a longest trail and show it cannot be improved.