Assuming the second condition is $A \cap B= \emptyset$. Consider $mn$ different primes $p_{i,j}$ where $ 1\leq i \leq m$ and $1\leq j \leq n$. Now, we consider $n$ consecutive positive integers $N+1,N+2,\dots N+n$ and $m$ consecutive integers $M+1,M+2,\dots M+m$. By CRT we can construct $N$ and $M$ such that for all $i$ we have $N+1 \equiv 0 \pmod {p_{i,1}}$, $N+2 \equiv 0 \pmod {p_{i,2}}, \dots, N+n \equiv 0 \pmod {p_{i,n}}$ and for all $j$ we have $M+1 \equiv 0 \pmod {p_{1,j}}$, $M+2 \equiv 0 \pmod {p_{2,j}}, \dots, M+m \equiv 0 \pmod {p_{m,j}}$. Clearly, for every pair $M+i$ and $N+j$ there exists a prime dividing both of them, and we can get these sets to be disjoint since infinitely many $M$ satisfy the given congruencies, and we can take $M> N+n$. $\blacksquare$