Solution: The first 12 even numbers after 2020 are $2022, 2024, \ldots, 2044$. The sum of an arithmetic sequence is $S = racn2(a_1 + a_n)$. Here, $n = 12$, $a_1 = 2022$, $a_n = 2044$. Thus, $S = rac122(2022 + 2044) = 6 imes 4066 = 24396$. Dividing 24396 by 11: $11 imes 2216 = 24376$, so the remainder is $24396 - 24376 = 20$. However, $20 \mod 11 = 9$. Therefore, the remainder is $oxed9$. - Nelissen Grade advocaten