2020 AMC 10A Problem 24

The first gcd condition gives $n+120\equiv0\pmod{21}$, so $n\equiv6\pmod{21}$, but $n+120$ must not be divisible by $63$. The second gives $n+63\equiv0\pmod{60}$, so $n\equiv57\pmod{60}$, but $n+63$ must not be divisible by $120$. Solving $n\equiv6\pmod{21}$ and $n\equiv57\pmod{60}$ gives $n\equiv237\pmod{420}$. The candidates above $1000$ are $1077,1497,1917,\ldots$. The first fails the first gcd condition, the second fails the second gcd condition, and $1917$ works. The digit sum is $18$. Thus, C is the correct answer.