Drawing K₈ without lifting the pen

Lower bound. Suppose the edges are covered by $s$ strokes. Count, for a fixed vertex $v$, the edges of $v$ covered by one stroke: if $v$ is not an endpoint of that stroke, each visit uses two edges at $v$, so the number is even. The total degree $7$ is odd, so some stroke uses an odd number of edges at $v$, and that stroke has $v$ as exactly one of its two ends. Each stroke has two ends, hence is such a stroke for at most $2$ vertices, while all $8$ vertices need one. Hence $2s\ge8$, $s\ge4$.

Construction with $4$ strokes. Pair the vertices into four pairs $\{1,2\},\{3,4\},\{5,6\},\{7,8\}$ and add a second, extra copy of each of these four edges. Now every vertex has degree $8$ (even) and the multigraph is connected, so it has an Eulerian circuit. Delete the four extra edges from the circuit: cutting a closed circuit at $4$ places leaves at most $4$ walks, which together use every original edge exactly once. So $4$ strokes suffice.

Answer. $4$ strokes.

Lesson. Parity of degrees gives the lower bound; to match it, fix the parity by adding edges and use Euler's theorem.