A geometric separator is a line that separates a given set of shapes to two subsets of approximately the same size (up to a constant), while intersecting only a small number of shapes. When a geometric separator exists, it is very useful because it allows us to solve difficult computational geometry problems in a divide-and-conquer manner.

Example: Given a set of $n$ disjoint axis-parallel squares in the plane, there is a rectangle such that at most $2n/3$ squares are inside it, at most $2n/3$ are outside it, and at most $O(\sqrt{n})$ are intersected by it.

Smith and Wormald (1998) prove this theorem, as well as thousands of generalizations with various applications. However, there is one generalization I haven't found yet, and I really want to know whether it is true:

**Given $m$ sets, each with $n$ disjoint axis-parallel squares of various sizes, is there a rectangle such that, at most $2mn/3$ squares are inside it, at most $2mn/3$ are outside it, and at most $O(\sqrt{n})$ squares of each collection are intersected by it?**

Note that, because the shapes in each of the $m$ sets are disjoint, the union of all sets is $m$-thick. Therefore, by theorem 39 in the original paper, it is possible to partition the collections such that the *total* number of intersected squares is $O(m\sqrt{n})$. Therefore, when $m=O(1)$, the existence of a simultaneous separator is guaranteed.

On the other hand, when $m \to \infty$, a simultaneous separator may not exist - see the comment by fedja.

The interesting case is when $m=\Omega(\sqrt{n})$. In this case, the existing theorem allows $\Omega(n)$ intersected squares in a single collection. My question is, basically, whether it is possible to bound the number of intersected squares *per collection*.