2006 AMC 10A Problem 9

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 $1 + 2 + 3 + 4 + 5.$ No set of $4$ works (their sum is even), and $6$ or more consecutive positive integers already exceed $15.$ There are $3$ such sets. Thus, the correct answer is C.