2011 AMC 10B Problem 21

The largest difference is $9$, so $w-z=9$. For either middle number $n$, the two differences $w-n$ and $n-z$ must add to $9$. The available pairs of differences that add to $9$ are $3+6$ and $4+5$, and the remaining difference between the two middle numbers is $1$. One possible set is $\{w,w-5,w-6,w-9\}$, giving $4w-20=44$ and $w=16$. The other is $\{w,w-3,w-4,w-9\}$, giving $4w-16=44$ and $w=15$. The sum of the possible values of $w$ is $16+15=31$. Thus, B is the correct answer.