Five boys and six girls are to be seated in a row of eleven chairs so that they sit one at a time from one end to the other. The probability that there are no more boys than girls seated at any point during the process is $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Evaluate $m + n$.
1. **Understanding the Problem:**
   We need to find the probability that there are no more boys than girls seated at any point during the process of seating 5 boys and 6 girls in a row of 11 chairs. This can be visualized as a path on a grid from \((0,0)\) to \((6,5)\) without crossing the line \(y = x\).

2. **Catalan Number:**
   The problem can be translated into counting the number of valid paths from \((0,0)\) to \((6,5)\) without crossing the line \(y = x\). This is a classic problem that can be solved using Catalan numbers. The \(n\)-th Catalan number \(C_n\) is given by:
   \[
   C_n = \frac{1}{n+1} \binom{2n}{n}
   \]
   For our problem, we need the 6th Catalan number \(C_6\):
   \[
   C_6 = \frac{1}{6+1} \binom{12}{6} = \frac{1}{7} \binom{12}{6}
   \]

3. **Calculating \(C_6\):**
   \[
   \binom{12}{6} = \frac{12!}{6!6!} = \frac{479001600}{720 \times 720} = 924
   \]
   Therefore,
   \[
   C_6 = \frac{1}{7} \times 924 = 132
   \]

4. **Total Number of Arrangements:**
   The total number of ways to arrange 5 boys and 6 girls in 11 chairs is given by:
   \[
   \binom{11}{5} = \frac{11!}{5!6!} = \frac{39916800}{120 \times 720} = 462
   \]

5. **Probability Calculation:**
   The probability that there are no more boys than girls seated at any point is:
   \[
   \frac{C_6}{\binom{11}{5}} = \frac{132}{462} = \frac{2}{7}
   \]

6. **Final Answer:**
   The fraction \(\frac{2}{7}\) is already in its simplest form, so \(m = 2\) and \(n = 7\). Therefore, \(m + n = 2 + 7 = 9\).

The final answer is \( \boxed{ 9 } \)