Let $A_1<A_2<A_3<\cdots <A_{2000}$ be the elements of $A$ and $B_1<B_2<\cdots <B_{2016}$ be the elements of $B$. Now we transform $A$ and $B$ in several steps as follows: $\to$ If in some stage, $A_1\le B_1$, then take the largest $p$ such that $A_1,A_2,\cdots A_p$ form a block of consecutive integers. Then take $A'=\{A_i+1|1\le i\le p\}\cup \{A_i|p<i\le 2000\}$. $B'= B$ in this case. $\to$ If $A_1>B_1$ instead, then take the largest $p$ such that $B_1,B_2,\cdots B_p$ form a block of consecutive integers. Then $B'=\{B_i+1|1\le i\le p\}\cup \{B_i|p<i\le 2016\}$. $A'=A$ in this case. After each step, $A'$ and $B'$ become $A$ and $B$ for the next step. It is easily verified that $K$ never decreases at each step. It is also obvious that after finitely many steps, each of $A$ and $B$ will consist of consecutive numbers and satisfy $A_1=B_1$. Since only differences in values matter, we can take WLOG $A_1=B_1=1$. In that case $A=\{1,2,...,2000\}$ and $B=\{1,2,...,2016\}$. It can be easily calculated that $K=3016880$ in this case. Therefore this is the required maximum.