Find $100m+n$ if $m$ and $n$ are relatively prime positive integers such that \[ \sum_{\substack{i,j \ge 0 \\ i+j \text{ odd}}} \frac{1}{2^i3^j} = \frac{m}{n}. \][i]Proposed by Aaron Lin[/i]
To solve the problem, we need to evaluate the sum 

\[
\sum_{\substack{i,j \ge 0 \\ i+j \text{ odd}}} \frac{1}{2^i3^j}
\]

and express it in the form \(\frac{m}{n}\) where \(m\) and \(n\) are relatively prime positive integers. We will then find \(100m + n\).

1. **Separate the sum based on the parity of \(i\) and \(j\):**
   - If \(i\) is odd and \(j\) is even.
   - If \(i\) is even and \(j\) is odd.

2. **Evaluate the sum for \(i\) odd and \(j\) even:**
   - For \(i\) odd, \(i = 1, 3, 5, \ldots\):
     \[
     \sum_{k=0}^{\infty} \frac{1}{2^{2k+1}} = \frac{1}{2} + \frac{1}{8} + \frac{1}{32} + \cdots
     \]
     This is a geometric series with the first term \(a = \frac{1}{2}\) and common ratio \(r = \frac{1}{4}\):
     \[
     \sum_{k=0}^{\infty} \frac{1}{2^{2k+1}} = \frac{\frac{1}{2}}{1 - \frac{1}{4}} = \frac{\frac{1}{2}}{\frac{3}{4}} = \frac{2}{3}
     \]
   - For \(j\) even, \(j = 0, 2, 4, \ldots\):
     \[
     \sum_{k=0}^{\infty} \frac{1}{3^{2k}} = 1 + \frac{1}{9} + \frac{1}{81} + \cdots
     \]
     This is a geometric series with the first term \(a = 1\) and common ratio \(r = \frac{1}{9}\):
     \[
     \sum_{k=0}^{\infty} \frac{1}{3^{2k}} = \frac{1}{1 - \frac{1}{9}} = \frac{1}{\frac{8}{9}} = \frac{9}{8}
     \]
   - Therefore, the product for \(i\) odd and \(j\) even is:
     \[
     \frac{2}{3} \cdot \frac{9}{8} = \frac{18}{24} = \frac{3}{4}
     \]

3. **Evaluate the sum for \(i\) even and \(j\) odd:**
   - For \(i\) even, \(i = 0, 2, 4, \ldots\):
     \[
     \sum_{k=0}^{\infty} \frac{1}{2^{2k}} = 1 + \frac{1}{4} + \frac{1}{16} + \cdots
     \]
     This is a geometric series with the first term \(a = 1\) and common ratio \(r = \frac{1}{4}\):
     \[
     \sum_{k=0}^{\infty} \frac{1}{2^{2k}} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}
     \]
   - For \(j\) odd, \(j = 1, 3, 5, \ldots\):
     \[
     \sum_{k=0}^{\infty} \frac{1}{3^{2k+1}} = \frac{1}{3} + \frac{1}{27} + \frac{1}{243} + \cdots
     \]
     This is a geometric series with the first term \(a = \frac{1}{3}\) and common ratio \(r = \frac{1}{9}\):
     \[
     \sum_{k=0}^{\infty} \frac{1}{3^{2k+1}} = \frac{\frac{1}{3}}{1 - \frac{1}{9}} = \frac{\frac{1}{3}}{\frac{8}{9}} = \frac{1}{3} \cdot \frac{9}{8} = \frac{3}{8}
     \]
   - Therefore, the product for \(i\) even and \(j\) odd is:
     \[
     \frac{4}{3} \cdot \frac{3}{8} = \frac{12}{24} = \frac{1}{2}
     \]

4. **Combine the results:**
   - The total sum is the sum of the two cases:
     \[
     \frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4}
     \]

Thus, we have \(\frac{m}{n} = \frac{5}{4}\) where \(m = 5\) and \(n = 4\). Therefore, \(100m + n = 100 \cdot 5 + 4 = 504\).

The final answer is \(\boxed{504}\).