How many integers must one select from the set $\{1,2, \ldots, 20\}$ to ensure that this selection includes two integers $a$ and $\mathrm{b}$ such that $a-b=2$?
Consider the drawers of the form $\{1,3\},\{2,4\},\{5,7\},\{6,8\},\{9,11\},\{10,12\},\{13,15\}$, $\{14,16\},\{17,19\},\{18,20\}$. By choosing 11 integers, by the pigeonhole principle, there will be two that are in the same drawer, and thus have a difference of 2. This quantity is minimal because the set $\{1,2,5,6,9,10,13,14,17,18\}$ is of size 10 and does not contain such integers $a$ and $b$.