2016 AMC 12B Problem 16

A run of $k$ consecutive integers equals $k$ times its median, and $345=3\cdot5\cdot23$

For odd $k,$ the median is an integer factor; for even $k,$ the median is a half-integer. Count both while keeping all terms positive

A sum of consecutive integers equals the count times the median. For an odd number of terms, the median is an integer divisor of $345,$ giving runs of $3$ (median $115$), $5$ (median $69$), $15$ (median $23$), and $23$ (median $15$) terms. For an even number of terms the median is a half-integer. The positive possibilities have lengths $2,6,10,$ with respective medians $172.5,57.5,34.5.$ Longer divisor-based runs would force a nonpositive first term. This gives $4+3=7$ ways.

Thus, the correct answer is E.