2024 AMC 10A Problem 4

Each two-digit number is at most $99,$ so $k$ of them sum to at most $99k.$ We need $99k \ge 2024,$ which forces $k \ge 20.4,$ so $k \ge 21.$ And $21$ really works: twenty $99$s plus one $44$ give $1980 + 44 = 2024.$ Therefore, the answer is B.