Let $S$ be a finite set, $I$ a finite index set, $\mathcal A=(A_i:i\in I)$ and $\mathcal B=(B_i:i\in I)$ families of subsets of $S$.
For $J\subseteq I$, let $A(J)$ denote $\bigcup_{j\in J} A_j$.
A partial transversal of $\mathcal A$ is the image of an injective map $f:J\to S$ from a subset $J$ of $I$ such that $f(i)\in A_i$ for all $i\in J$. We have a transversal if $J=I$. A common partial transversal of $\mathcal A$ and $\mathcal B$ is a subset that is a partial transversal of $\mathcal A$ and $\mathcal B$; it is a common transversal if the subset is a transversal of $\mathcal A$ and $\mathcal B$.
The following is the finite version of Theorem 7 in the article by J. de Sousa (also publishing as Joan Davies), "Disjoint Common Transversals," Combinatorial Mathematics and its Applications: Proceedings of a Conference held at the Mathematical Institute, Oxford, from 7-10 July, 1969 (D.J.A. Welsh, ed.).
If $m\ge1$, then $\mathcal A$ and $\mathcal B$ have $m$ pairwise disjoint common transversals if and only if, for all $J,K\subseteq I$, $|A(J)\cap B(K)|\ge m(|J|+|K|-|I|)$.
For $\\\$$123 (I am not Lavinia Clay! and please see the caveat below), can someone provide necessary and sufficient conditions for $\mathcal A$ and $\mathcal B$ to have $m\in\mathbb N$ pairwise disjoint common partial transversals of cardinalities $k_1$, $k_2$, $\dots$, $k_m\in\mathbb N$, under the assumption that the sets in $\mathcal A$ are pairwise disjoint, and likewise for $\mathcal B$?
The caveat is that this offer must be legal and allowed by the moderators; there must be some reasonable way for me to pay you (e.g., you are not in the Witness Protection Program or North Korea); I am the sole arbiter of what is an acceptable solution; there can be only one winner; I reserve the right to change my mind in case I missed some condition; and I place a time limit of 3 months from today.
See Theorem 2.2 of D. R. Fulkerson, "Disjoint Common Partial Transversals of Two Families of Sets," RAND Memorandum, Document RM-6102-PR (1969).
