Note that the measurable lengths must be the lengths from $1$ to $l$ inclusive. There are $\binom{k+1}{2}$ distinct measurable distances, because the distances between any two marks are pairwise distinct. Thus $l = \binom{k+1}{2}$. For $k=1, 2, 3$, standard case work can be used to find all Golomb rulers. We obtain: $l=1$, marks are $0, 1$; $l=3$, marks are $0, 1, 3$ or $0, 2, 3$; $l=6$, marks are $0, 1, 4, 6$ or $0, 2, 5, 6$. Now assume that $k \ge 4$. For $\binom{k+1}{2}-1$ to be a measurable distance, $1$ or $\binom{k+1}{2}-1$ must be marked, wlog the former. For $\binom{k+1}{2}-2$ to be a measurable length, one of the pairs $(0, \binom{k+1}{2}-2)$, $(1, \binom{k+1}{2}-1)$, $(2, \binom{k+1}{2})$ must have both points marked. In the second pair, the distance $1$ can be measured twice, and in the third pair, the distance $1$ can also be measured twice. Thus the first pair must occur, implying that $\binom{k+1}{2}-2$ is marked. For $\binom{k+1}{2} - 4$ to be a measurable distance, one of the pairs $(0, \binom{k+1}{2}-4), (1, \binom{k+1}{2}-3), (2, \binom{k+1}{2}-2), (3, \binom{k+1}{2}-1), (\binom{k+1}{2})$ must have both points marked. In the first case, the distance $2$ can be measured twice, in the second case, the distance $1$ can be measured twice, same in the third case. Thus $4$ must be marked. Now we can go through the possibilities for which $\binom{k+1}{2} - 5$ can be measured, and we find that in each case, there is a distance that can be measured twice. Thus there are no Golomb rulers for $k \ge 4$.