How to reduce the original partition problem to one of its variation?
Thanks to D.W.'s generous help, I managed to figure it out. Given an instance of the Partition problem, $S$, we can just scale every element in $S$ by 1000, and the new $S'$ becomes an instance of the variation. Then the reduction mapping holds, because scaling increases the difference between numbers. If $S$ is a YES instance, $S'$ can also be partitioned into two sets with equal sum, but if $S$ is a NO instance, all of the possible partitions of $S'$ must have difference $\geq 1000$. The reduction is clearly polynomial, and the variation is clearly in NP, so proof is complete.