The sides $x$ and $y$ of a scalene triangle satisfy $x + \frac{2\Delta }{x}=y+ \frac{2\Delta }{y}$ , where $\Delta$ is the area of the triangle. If $x = 60, y = 63$, what is the length of the largest side of the triangle?
1. Given the equation for the sides of the triangle:
   \[
   x + \frac{2\Delta}{x} = y + \frac{2\Delta}{y} = k
   \]
   where \( \Delta \) is the area of the triangle, and \( x = 60 \), \( y = 63 \).

2. We can rewrite the given equation as:
   \[
   x + \frac{2\Delta}{x} = k \quad \text{and} \quad y + \frac{2\Delta}{y} = k
   \]

3. Rearrange these equations to express \( k \):
   \[
   k = x + \frac{2\Delta}{x} \quad \text{and} \quad k = y + \frac{2\Delta}{y}
   \]

4. Since both expressions equal \( k \), we can set them equal to each other:
   \[
   x + \frac{2\Delta}{x} = y + \frac{2\Delta}{y}
   \]

5. Substitute \( x = 60 \) and \( y = 63 \):
   \[
   60 + \frac{2\Delta}{60} = 63 + \frac{2\Delta}{63}
   \]

6. Simplify the fractions:
   \[
   60 + \frac{\Delta}{30} = 63 + \frac{2\Delta}{63}
   \]

7. Multiply through by 63 to clear the denominators:
   \[
   60 \cdot 63 + \frac{63\Delta}{30} = 63 \cdot 63 + 2\Delta
   \]

8. Simplify the equation:
   \[
   3780 + \frac{63\Delta}{30} = 3969 + 2\Delta
   \]

9. Multiply through by 30 to clear the fraction:
   \[
   113400 + 63\Delta = 119070 + 60\Delta
   \]

10. Rearrange to solve for \( \Delta \):
    \[
    113400 + 63\Delta - 60\Delta = 119070
    \]
    \[
    113400 + 3\Delta = 119070
    \]
    \[
    3\Delta = 119070 - 113400
    \]
    \[
    3\Delta = 5670
    \]
    \[
    \Delta = 1890
    \]

11. Now, we use the fact that \( x \) and \( y \) are the roots of the quadratic equation:
    \[
    z^2 - kz + 2\Delta = 0
    \]

12. By Vieta's formulas, the product of the roots \( x \) and \( y \) is given by:
    \[
    xy = 2\Delta
    \]
    \[
    60 \cdot 63 = 2 \cdot 1890
    \]
    \[
    3780 = 3780
    \]

13. Since the product of the roots confirms the area, we conclude that the triangle is a right triangle with sides \( x = 60 \) and \( y = 63 \).

14. The hypotenuse \( c \) of the right triangle can be found using the Pythagorean theorem:
    \[
    c = \sqrt{x^2 + y^2}
    \]
    \[
    c = \sqrt{60^2 + 63^2}
    \]
    \[
    c = \sqrt{3600 + 3969}
    \]
    \[
    c = \sqrt{7569}
    \]
    \[
    c = 87
    \]

The final answer is \(\boxed{87}\).