a is an element of A and b is an element of B, there are 3998 values of a+b also there are at least 3999 pair of a,b so there are four numbers a,b,c,d such that a and c are elements of A; b and d are elements of B and a+b=c+d hence a-c=b-d gives the desired result.

Consider the function $F: A \times B \rightarrow \{2,3,....,4000\}$ defined by $F(a,b)= a+b$. Let us consider two cases: 1. $F$ is onto. Then both $A$ nad $B$ contain $1$ and $2000$. Therefore, $1999= 2000-1$ belongs to $(A-A) \cap (B-B)$. 2. $F$ is not onto. Since $|A| \times |B| \geq 3999$, it follows by the pigeonhole prinbiple that $F$ cannot be injective (since $F$ may have a maximum of $3999$ possible values and $F$ does not necessarily assume all of them). So, we must have $a_{1},a_{2}$ in $A$ and $b_{1},b_{2}$ in $B$ such that $a_{1}+b_{1}=a_{2}+b_{2}$. From this we get $a_{1}-a_{2}=b_{2}-b_{1}$, and since $a_{1}-a_{2}\in A-A$ and $b_{2}-b_{1}\in B-B$, it follows that $A-A \cap B-B$ is non-empty, which completes the proof.