2005 AMC 10B Problem 25

The pairs summing to $125$ are $(25, 100), (26, 99), \ldots, (62, 63),$ which is $62 - 25 + 1 = 38$ pairs. From each pair, $B$ may contain at most one element. The numbers $1$ through $24$ cannot pair with anything in range to sum to $125,$ so all $24$ of them may be included. Thus $B$ has at most $38 + 24 = 62$ elements, and the set $\{1, 2, \ldots, 62\}$ achieves this. Thus, C is the correct answer.