2024 AMC 12B Problem 5

The original expression sums the first $50$ odd numbers, giving $50^2 = 2500.$ Changing a term of value $v$ from $+$ to $-$ decreases the total by $2v,$ so to make the result negative the flipped terms must total more than $\dfrac{2500}{2} = 1250.$ To use as few terms as possible, flip the largest odd numbers $99, 97, 95, \ldots$ The largest $k$ of them sum to $k(100 - k).$ With $k = 14$ this is $14 \cdot 86 = 1204 \le 1250,$ but with $k = 15$ it is $15 \cdot 85 = 1275 \gt 1250.$ So $15$ sign changes suffice and $14$ do not. Thus, the correct answer is B.