Express these $100$ numbers as $2^k \cdot j$, where $j=1$, $3$, $5$, $\dots$, $99$, and then group all the numbers by $j$: {1, 2, 4, 8, 16, 32, 64} {3, 6, 12, 24, 48,96} {5, 10, 20, 40, 80} ... {99} There are totally $50$ such sets. By the pigeonhole principle, at least two chosen numbers will be in the same set. By construction of these sets, these two numbers must meet the requirement.