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 $
1. We start with the given conditions:
   \[
   w + x + y + z = 44
   \]
   \[
   w > x > y > z
   \]
   The pairwise positive differences are \(1, 3, 4, 5, 6,\) and \(9\).

2. Since \(w > x > y > z\), the largest difference must be \(w - z = 9\).

3. We need to consider the smallest difference, which must be \(1\). This difference can be \(w - x\), \(x - y\), or \(y - z\).

4. **Case 1: \(w - x = 1\)**
   \[
   w - z = 9 \quad \text{and} \quad w - x = 1
   \]
   This implies:
   \[
   (w - z) - (w - x) = 9 - 1 = 8 \quad \Rightarrow \quad x - z = 8
   \]
   However, \(x - z = 8\) is not in the list of differences. Therefore, this case fails.

5. **Case 2: \(x - y = 1\)**
   \[
   w - z = 9 \quad \text{and} \quad x - y = 1
   \]
   This implies:
   \[
   (w - z) + (x - y) = 9 + 1 = 10
   \]
   Combining with the sum of the integers:
   \[
   w + x + y + z = 44
   \]
   We can rewrite \(w + x + y + z\) as:
   \[
   w + x + y + z = (w - z) + (x - y) + 2y + 2z = 10 + 2y + 2z = 44
   \]
   Simplifying, we get:
   \[
   2y + 2z = 34 \quad \Rightarrow \quad y + z = 17
   \]
   Since \(y + z = 17\) and \(y - z\) must be one of \(3, 4, 5, 6\), and the sum of two integers is odd, their difference must be odd. Thus, \(y - z = 3\) or \(y - z = 5\).

   - If \(y - z = 3\):
     \[
     y + z = 17 \quad \text{and} \quad y - z = 3
     \]
     Solving these equations:
     \[
     2y = 20 \quad \Rightarrow \quad y = 10
     \]
     \[
     2z = 14 \quad \Rightarrow \quad z = 7
     \]
     Thus, \((y, z) = (10, 7)\).

     Using \(w - z = 9\):
     \[
     w = z + 9 = 7 + 9 = 16
     \]
     Using \(x - y = 1\):
     \[
     x = y + 1 = 10 + 1 = 11
     \]
     Therefore, \((w, x, y, z) = (16, 11, 10, 7)\).

   - If \(y - z = 5\):
     \[
     y + z = 17 \quad \text{and} \quad y - z = 5
     \]
     Solving these equations:
     \[
     2y = 22 \quad \Rightarrow \quad y = 11
     \]
     \[
     2z = 12 \quad \Rightarrow \quad z = 6
     \]
     Thus, \((y, z) = (11, 6)\).

     Using \(w - z = 9\):
     \[
     w = z + 9 = 6 + 9 = 15
     \]
     Using \(x - y = 1\):
     \[
     x = y + 1 = 11 + 1 = 12
     \]
     Therefore, \((w, x, y, z) = (15, 12, 11, 6)\).

6. **Case 3: \(y - z = 1\)**
   \[
   w - z = 9 \quad \text{and} \quad y - z = 1
   \]
   This implies:
   \[
   (w - z) - (y - z) = 9 - 1 = 8 \quad \Rightarrow \quad w - y = 8
   \]
   However, \(w - y = 8\) is not in the list of differences. Therefore, this case fails.

7. Summarizing the valid cases, we have:
   \[
   (w, x, y, z) = (16, 11, 10, 7) \quad \text{and} \quad (w, x, y, z) = (15, 12, 11, 6)
   \]
   The possible values for \(w\) are \(16\) and \(15\).

8. The sum of the possible values for \(w\) is:
   \[
   16 + 15 = 31
   \]

The final answer is \(\boxed{31}\)