Brian writes down four integers $w > x > y > z$ whose sum is $44$. The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$. What is the sum of the possible values for $w$?
$\textbf{(A)}\ 16 \qquad\textbf{(B)}\ 31 \qquad\textbf{(C)}\ 48 \qquad\textbf{(D)}\ 62 \qquad\textbf{(E)}\ 93$
The largest difference, $9,$ must be between $w$ and $z.$
The smallest difference, $1,$ must be directly between two integers. This also means the differences directly between the other two should add up to $8.$ The only remaining differences that would make this possible are $3$ and $5.$ However, those two differences can't be right next to each other because they would make a difference of $8,$ which isn't given as a possibility in the problem. This means $1$ must be the difference between $y$ and $x.$ We can express the possible configurations as the lines.




If we look at the first number line, you can express $x$ as $w-5,$ $y$ as $w-6,$ and $z$ as $w-9.$ Since the sum of all these integers equal $44$,
\begin{align*} w+w-5+w-6+w-9&=44\\ 4w&=64\\ w&=16 \end{align*}
You can do something similar to this with the second number line to find the other possible value of $w.$
\begin{align*} w+w-3+w-4+w-9&=44\\ 4w&=60\\ w&=15 \end{align*}
The sum of the possible values of $w$ is $16+15 = \boxed{\textbf{(B) }31}$