 Solution

Take the nonempty intersections of the length-three and length-four blocks. Within the first twelve nonnegative integers the common cut points are $0,3,4,6,8,9,12$, givingThe complete partition of a set consists of these six blocks translated by $12j$ for every $j\in \mathbb{N}$. Indeed, adding $12$ increases the two floor quotients by $4$ and $3$ respectively, preserving equality of both quotients within each translated block. No adjacent displayed blocks can merge because at least one quotient changes at their boundary. This is the periodic common refinement of floor-quotient partitions, with period $\mathrm{lcm}(3,4)=12$.