10.1. Let these numbers be A and B. Let B be the smallest of these numbers. Then the larger number $\mathrm{A}=10 \mathrm{~B}+\mathrm{n}$, where $\mathrm{n}$ is the crossed-out digit, and

$0 \leq \mathrm{n} \leq 9$. According to the condition, $\mathrm{A}+\mathrm{B}=11 \mathrm{~B}+\mathrm{n}=2022$. Therefore, $\mathrm{n}$ is the remainder of the division of 2022 by 11, which gives $\mathrm{n}=9$. We get $\mathrm{B}=183, \mathrm{~A}=1839$.