There are some weights on the two sides of a balance scale. The mass of each weight is an integer number of grams, but no two weights on the same side of the scale share the same mass. At the moment, the scale is perfectly balanced, with each side weighing a total of W grams. Suppose W is less than the number of weights on the left multiplied by the number of weights on the right. Is it always true that we can remove some of the weights from each side and still keep the two sides balanced?
Rem: Ignore the trivial solution of removing all weights from both sides.
Yes, because there must a mass shared by two weights on different sides.
Suppose for contradiction that there is no such mass. Then, the weights are all distinct. Let $a$ and $b$ be the number of weights on the two sides. At minimum, the total weight of the two sides is $(a+b)(a+b+1)/2$ when the masses are $1,2,\dots,a+b$.
Then, we have
$$2w \geq(a+b)(a+b+1)/2 > (a+b)^2/2 \geq 2ab $$
where the last inequality follows from being equivalent to $(a-b)^2/2 \geq 0$. But this contradicts the statement that $w<ab$.