Pentagon $ANDD'Y$ has $AN \parallel DY$ and $AY \parallel D'N$ with $AN = D'Y$ and $AY = DN$. If the area of $ANDY$ is 20, the area of $AND'Y$ is 24, and the area of $ADD'$ is 26, the area of $ANDD'Y$ can be expressed in the form $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Find $m+n$.

[i]Proposed by Andy Xu[/i]
Given the pentagon \(ANDD'Y\) with the following properties:
- \(AN \parallel DY\)
- \(AY \parallel D'N\)
- \(AN = D'Y\)
- \(AY = DN\)

We are also given the areas of certain regions:
- The area of \(ANDY\) is 20.
- The area of \(AND'Y\) is 24.
- The area of \(ADD'\) is 26.

We need to find the area of the pentagon \(ANDD'Y\) and express it in the form \(\frac{m}{n}\) for relatively prime positive integers \(m\) and \(n\), and then find \(m+n\).

1. **Identify the areas of the trapezoids and triangles:**
   - The trapezoid \(ANDY\) has an area of 20.
   - The trapezoid \(AND'Y\) has an area of 24.
   - The triangle \(ADD'\) has an area of 26.

2. **Use the properties of the trapezoids:**
   Since \(AN \parallel DY\) and \(AY \parallel D'N\), both \(ANDY\) and \(AND'Y\) are isosceles trapezoids. This implies that the opposite sides are equal and the heights from the parallel sides to the bases are the same.

3. **Calculate the total area of the pentagon \(ANDD'Y\):**
   The pentagon \(ANDD'Y\) can be divided into the trapezoid \(ANDY\), the trapezoid \(AND'Y\), and the triangle \(ADD'\).

   The total area of the pentagon is the sum of the areas of these three regions:
   \[
   \text{Area of } ANDD'Y = \text{Area of } ANDY + \text{Area of } AND'Y + \text{Area of } ADD'
   \]
   Substituting the given areas:
   \[
   \text{Area of } ANDD'Y = 20 + 24 + 26 = 70
   \]

4. **Express the area in the form \(\frac{m}{n}\):**
   The area of the pentagon \(ANDD'Y\) is already a whole number, 70, which can be expressed as \(\frac{70}{1}\).

5. **Find \(m+n\):**
   Here, \(m = 70\) and \(n = 1\). Therefore, \(m+n = 70 + 1 = 71\).

The final answer is \( \boxed{ 71 } \).