We call a pair two weights whose sum is $201\ \mathrm{g}$. We wish to obtain, in the end, $50$ pairs on each of the two pans of the balance.

If on the pan on the left we have the weights $a_1, a_2, \dots, a_{50}$ and their pairs $b_1, b_2, \dots, b_{50}$ are on the pan on the right, we move the weights such that, in the end, on the left pan we have the weights $a_1, a_2, \dots, a_{50}$ together with their pairs, $b_1, b_2, \dots, b_{50}$.

If we have less than $50$ pairs that are split between the two pans, we must have at least $25$ pairs on the left pan and (at least) $25$ pairs on the right pan. Moving $25$ complete pairs from the right pan next to $25$ pairs from the left pan, we obtain, again, $50$ complete pairs on one pan, hence the desired result.