= Solution
Take the nonempty intersections of the length-three and length-four blocks. Within the first twelve <nonnegative integers> the common cut points are $0,3,4,6,8,9,12$, giving
$$
\{0,1,2\},\quad\{3\},\quad\{4,5\},\quad\{6,7\},\quad\{8\},\quad\{9,10,11\}.
$$
The 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 $\operatorname{lcm}(3,4)=12$.
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