Let $ a$ and $ b$ be positive real numbers with $ a\ge b$. Let $ \rho$ be the maximum possible value of $ \frac{a}{b}$ for which the system of equations
\[ a^2\plus{}y^2\equal{}b^2\plus{}x^2\equal{}(a\minus{}x)^2\plus{}(b\minus{}y)^2\]has a solution in $ (x,y)$ satisfying $ 0\le x<a$ and $ 0\le y<b$. Then $ \rho^2$ can be expressed as a fraction $ \frac{m}{n}$, where $ m$ and $ n$ are relatively prime positive integers. Find $ m\plus{}n$.
1. Consider the given system of equations:
   \[
   a^2 + y^2 = b^2 + x^2 = (a - x)^2 + (b - y)^2
   \]
   We need to find the maximum possible value of \(\frac{a}{b}\) for which this system has a solution \((x, y)\) satisfying \(0 \le x < a\) and \(0 \le y < b\).

2. Let's analyze the geometric interpretation of the equations. The equations suggest that the points \((a, y)\), \((x, b)\), and \((a - x, b - y)\) form an equilateral triangle. This is because the distances between these points are equal.

3. Consider the points \(O(0,0)\), \(A(a, y)\), and \(B(x, b)\). The problem statement implies that these points form an equilateral triangle. 

4. In an equilateral triangle, the ratio of the side length to the height is \(\frac{2}{\sqrt{3}}\). This is derived from the properties of an equilateral triangle where the height \(h\) of a triangle with side length \(s\) is given by:
   \[
   h = \frac{\sqrt{3}}{2} s
   \]
   Therefore, the ratio of the side length to the height is:
   \[
   \frac{s}{h} = \frac{s}{\frac{\sqrt{3}}{2} s} = \frac{2}{\sqrt{3}}
   \]

5. Given that \(a \ge b\), the maximum possible value of \(\frac{a}{b}\) is \(\frac{2}{\sqrt{3}}\).

6. To express \(\rho^2\) as a fraction \(\frac{m}{n}\), we calculate:
   \[
   \rho = \frac{2}{\sqrt{3}} \implies \rho^2 = \left(\frac{2}{\sqrt{3}}\right)^2 = \frac{4}{3}
   \]
   Here, \(m = 4\) and \(n = 3\), which are relatively prime.

7. Therefore, \(m + n = 4 + 3 = 7\).

The final answer is \(\boxed{7}\).