 Solution

The fifty disjoint sets $\{1,100\},\{2,99\},\dots ,\{50,51\}$ form a partition of a set $\{1,\dots ,100\}$. If a chosen subset met each pair at most once, it would contain at most fifty elements. The pigeonhole principle therefore forces a $51$-element subset to contain both elements of at least one pair. Those two distinct elements sum to $101$.