2024 AMC 12A Problem 3

Each number is at most $99,$ so $k$ numbers sum to at most $99k.$ Since $99\cdot20=1980\lt2024,$ at least $21$ numbers are required. With $21,$ we can use twenty $99$s and one $44$: $20\cdot99+44=1980+44$ $=2024.$ Thus, the correct answer is B.