Daniel had a string that formed the perimeter of a square with area $98$. Daniel cut the string into two pieces. With one piece he formed the perimeter of a rectangle whose width and length are in the ratio $2 : 3$. With the other piece he formed the perimeter of a rectangle whose width and length are in the ratio $3 : 8$. The two rectangles that Daniel formed have the same area, and each of those areas is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
1. **Determine the perimeter of the original square:**
   - Given the area of the square is \(98\), we can find the side length \(s\) of the square using the formula for the area of a square:
     \[
     s^2 = 98 \implies s = \sqrt{98} = 7\sqrt{2}
     \]
   - The perimeter \(P\) of the square is:
     \[
     P = 4s = 4 \times 7\sqrt{2} = 28\sqrt{2}
     \]

2. **Set up the perimeters of the two rectangles:**
   - Let the perimeter of the first rectangle be \(10x\) (since the width and length are in the ratio \(2:3\)).
   - Let the perimeter of the second rectangle be \(22y\) (since the width and length are in the ratio \(3:8\)).
   - The total length of the string is the sum of the perimeters of the two rectangles:
     \[
     10x + 22y = 28\sqrt{2}
     \]

3. **Set up the areas of the two rectangles:**
   - For the first rectangle with width \(2x\) and length \(3x\), the area \(A_1\) is:
     \[
     A_1 = 2x \cdot 3x = 6x^2
     \]
   - For the second rectangle with width \(3y\) and length \(8y\), the area \(A_2\) is:
     \[
     A_2 = 3y \cdot 8y = 24y^2
     \]
   - Given that the areas of the two rectangles are equal:
     \[
     6x^2 = 24y^2 \implies x^2 = 4y^2 \implies x = 2y
     \]

4. **Solve for \(y\):**
   - Substitute \(x = 2y\) into the perimeter equation:
     \[
     10(2y) + 22y = 28\sqrt{2} \implies 20y + 22y = 28\sqrt{2} \implies 42y = 28\sqrt{2} \implies y = \frac{28\sqrt{2}}{42} = \frac{2\sqrt{2}}{3}
     \]

5. **Calculate the area of the rectangles:**
   - Substitute \(y = \frac{2\sqrt{2}}{3}\) into the area formula for the second rectangle:
     \[
     A_2 = 24y^2 = 24 \left(\frac{2\sqrt{2}}{3}\right)^2 = 24 \left(\frac{8}{9}\right) = \frac{192}{9} = \frac{64}{3}
     \]

6. **Find \(m + n\):**
   - The area of each rectangle is \(\frac{m}{n} = \frac{64}{3}\), where \(m = 64\) and \(n = 3\).
   - Therefore, \(m + n = 64 + 3 = 67\).

The final answer is \(\boxed{67}\)