In order to save money on gas and use up less fuel, Hannah has a special battery installed in the family van.  Before the installation, the van averaged $18$ miles per gallon of gas.  After the conversion, the van got $24$ miles per gallong of gas.

Michael notes, "The amount of money we will save on gas over any time period is equal to the amount we would save if we were able to convert the van to go from $24$ miles per gallon to $m$ miles per gallon.  It is also the same that we would save if we were able to convert the van to go from $m$ miles per gallon to $n$ miles per gallon."

Assuming Michael is correct, compute $m+n$.  In this problem, assume that gas mileage is constant over all speeds and terrain and that the van gets used the same amount regardless of its present state of conversion.
1. **Initial Setup:**
   - Before the conversion, the van averaged \(18\) miles per gallon (mpg).
   - After the conversion, the van averaged \(24\) mpg.

2. **Calculate the improvement in fuel efficiency:**
   - Before conversion: \(18\) mpg means \(6\) miles per \(\frac{6}{18} = \frac{1}{3}\) gallon.
   - After conversion: \(24\) mpg means \(6\) miles per \(\frac{6}{24} = \frac{1}{4}\) gallon.
   - The improvement in fuel efficiency is \(\frac{1}{3} - \frac{1}{4} = \frac{4}{12} - \frac{3}{12} = \frac{1}{12}\) gallon per \(6\) miles.

3. **Determine the value of \(m\):**
   - \(m\) mpg should improve the fuel efficiency by another \(\frac{1}{12}\) gallon per \(6\) miles.
   - If \(m\) mpg means \(6\) miles per \(\frac{6}{m}\) gallons, then:
     \[
     \frac{1}{4} - \frac{6}{m} = \frac{1}{12}
     \]
   - Solving for \(m\):
     \[
     \frac{1}{4} - \frac{6}{m} = \frac{1}{12}
     \]
     \[
     \frac{1}{4} - \frac{1}{12} = \frac{6}{m}
     \]
     \[
     \frac{3}{12} - \frac{1}{12} = \frac{6}{m}
     \]
     \[
     \frac{2}{12} = \frac{6}{m}
     \]
     \[
     \frac{1}{6} = \frac{6}{m}
     \]
     \[
     m = 36
     \]

4. **Determine the value of \(n\):**
   - \(n\) mpg should improve the fuel efficiency by another \(\frac{1}{12}\) gallon per \(6\) miles.
   - If \(n\) mpg means \(6\) miles per \(\frac{6}{n}\) gallons, then:
     \[
     \frac{6}{m} - \frac{6}{n} = \frac{1}{12}
     \]
   - Since \(m = 36\):
     \[
     \frac{6}{36} - \frac{6}{n} = \frac{1}{12}
     \]
     \[
     \frac{1}{6} - \frac{6}{n} = \frac{1}{12}
     \]
     \[
     \frac{1}{6} - \frac{1}{12} = \frac{6}{n}
     \]
     \[
     \frac{2}{12} - \frac{1}{12} = \frac{6}{n}
     \]
     \[
     \frac{1}{12} = \frac{6}{n}
     \]
     \[
     n = 72
     \]

5. **Sum of \(m\) and \(n\):**
   \[
   m + n = 36 + 72 = 108
   \]

The final answer is \(\boxed{108}\)