, 2013

Solution:
Answer: $\binom{4025}{2012}$

In general, we will show that if each of the sets $A_{i}$ contain $a$ elements and if each of the sets $B_{j}$ contain $b$ elements, then the maximum value for $m$ is $\binom{a+b}{a}$.

Let $U$ denote the union of all the sets $A_{i}$ and $B_{j}$ and let $|U| = n$. Consider the $n!$ orderings of the elements of $U$. Note that for any specific ordering, there is at most one value of $i$ such that all the elements in $A_{i}$ come before all the elements in $B_{i}$ in this ordering; this follows since $A_{j}$ shares at least one element with $B_{i}$ and $B_{j}$ shares at least one element with $A_{i}$ for any other $j \neq i$.

On the other hand, the number of ways to permute the $(a+b)$ elements in $A_{i} \cup B_{i}$ so that all the elements in $A_{i}$ come first is equal to $a!b!$. Therefore, the number of permutations of $U$ where all the elements in $A_{i}$ come before all the elements in $B_{i}$ is equal to:
$$n! \cdot \frac{a!b!}{(a+b)!} = \frac{n!}{\binom{a+b}{a}}$$
Summing over all $m$ values of $i$, the total number of orderings where, for some $i$, the elements in $A_{i}$ come before $B_{i}$ is equal to
$$\frac{n! m}{\binom{a+b}{a}}$$
But there are at most $n!$ such orderings, since there are $n!$ total orderings, so it follows that $m \leq \binom{a+b}{a}$. Equality is attained by taking $U$ to be a set containing $(a+b)$ elements, letting $A_{i}$ range over all $a$-element subsets of $U$, and letting $B_{i} = U \setminus A_{i}$ for each $i$.