Let $ABCD$ be a square of side length $4$. Points $E$ and $F$ are chosen on sides $BC$ and $DA$, respectively, such that $EF = 5$. Find the sum of the minimum and maximum possible areas of trapezoid $BEDF$.

[i]Proposed by Andrew Wu[/i]
1. **Define the problem and set up coordinates:**
   Let the square \(ABCD\) have vertices at \(A(0, 4)\), \(B(0, 0)\), \(C(4, 0)\), and \(D(4, 4)\). Points \(E\) and \(F\) are on sides \(BC\) and \(DA\) respectively, such that \(EF = 5\).

2. **Determine the coordinates of \(E\) and \(F\):**
   Let \(E\) be at \((4, y_E)\) on \(BC\) and \(F\) be at \((x_F, 4)\) on \(DA\). Since \(EF = 5\), we use the distance formula:
   \[
   \sqrt{(4 - x_F)^2 + (y_E - 4)^2} = 5
   \]
   Simplifying, we get:
   \[
   (4 - x_F)^2 + (y_E - 4)^2 = 25
   \]

3. **Analyze the vertical and horizontal distances:**
   Since \(E\) is on \(BC\) and \(F\) is on \(DA\), the horizontal distance between \(E\) and \(F\) is \(4\). Therefore, the vertical distance must be:
   \[
   \sqrt{25 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3
   \]

4. **Determine the possible positions of \(E\) and \(F\):**
   - **Case 1:** \(E\) is close to \(B\) and \(F\) is close to \(D\):
     - \(y_E = 3\) (since \(E\) is 3 units above \(B\))
     - \(x_F = 1\) (since \(F\) is 3 units to the left of \(D\))
   - **Case 2:** \(E\) is far from \(B\) and \(F\) is far from \(D\):
     - \(y_E = 1\) (since \(E\) is 1 unit above \(C\))
     - \(x_F = 3\) (since \(F\) is 1 unit to the right of \(A\))

5. **Calculate the areas of trapezoid \(BEDF\):**
   - **Case 1:**
     - Bases: \(BE = 3\) and \(DF = 1\)
     - Height: \(4\)
     - Area: 
       \[
       \text{Area}_1 = \frac{1}{2} \times (3 + 1) \times 4 = \frac{1}{2} \times 4 \times 4 = 8
       \]
   - **Case 2:**
     - Bases: \(BE = 1\) and \(DF = 3\)
     - Height: \(4\)
     - Area:
       \[
       \text{Area}_2 = \frac{1}{2} \times (1 + 3) \times 4 = \frac{1}{2} \times 4 \times 4 = 8
       \]

6. **Sum of the minimum and maximum possible areas:**
   Since both cases yield the same area, the sum of the minimum and maximum possible areas is:
   \[
   8 + 8 = 16
   \]

The final answer is \(\boxed{16}\)