2006 AMC 12A Problem 8

The sum of $n$ consecutive integers equals $n$ times their median. For a sum of $15:$ $n = 2$ gives $7 + 8,$ $n = 3$ gives $4 + 5 + 6,$ and $n = 5$ gives $1 + 2 + 3 + 4 + 5.$ A run of four consecutive integers sums to an even number, and more than five terms already exceed $1 + 2 + 3 + 4 + 5 = 15.$ So there are $3$ sets. Thus, the correct answer is C.