In January Jeff’s investment went up by three quarters. In February it went down by one quarter. In March it went up by one third. In April it went down by one fifth. In May it went up by one seventh. In June Jeff’s investment fell by $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. If Jeff’s investment was worth the same amount at the end of June as it had been at the beginning of January, find $m + n$.
1. Let Jeff's initial investment be $I = 1$ (for simplicity).
2. In January, the investment went up by three quarters:
   \[
   I_{\text{Jan}} = 1 + \frac{3}{4} = \frac{7}{4}
   \]
3. In February, the investment went down by one quarter:
   \[
   I_{\text{Feb}} = \frac{7}{4} \times \left(1 - \frac{1}{4}\right) = \frac{7}{4} \times \frac{3}{4} = \frac{21}{16}
   \]
4. In March, the investment went up by one third:
   \[
   I_{\text{Mar}} = \frac{21}{16} \times \left(1 + \frac{1}{3}\right) = \frac{21}{16} \times \frac{4}{3} = \frac{7}{4}
   \]
5. In April, the investment went down by one fifth:
   \[
   I_{\text{Apr}} = \frac{7}{4} \times \left(1 - \frac{1}{5}\right) = \frac{7}{4} \times \frac{4}{5} = \frac{7}{5}
   \]
6. In May, the investment went up by one seventh:
   \[
   I_{\text{May}} = \frac{7}{5} \times \left(1 + \frac{1}{7}\right) = \frac{7}{5} \times \frac{8}{7} = \frac{8}{5}
   \]
7. In June, the investment fell by $\frac{m}{n}$, and we know the investment at the end of June is the same as it was at the beginning of January:
   \[
   \frac{8}{5} \times \left(1 - \frac{m}{n}\right) = 1
   \]
8. Solving for $\frac{m}{n}$:
   \[
   \frac{8}{5} \times \left(1 - \frac{m}{n}\right) = 1
   \]
   \[
   1 - \frac{m}{n} = \frac{5}{8}
   \]
   \[
   \frac{m}{n} = 1 - \frac{5}{8} = \frac{3}{8}
   \]
9. Since $m$ and $n$ are relatively prime, we have $m = 3$ and $n = 8$.
10. Therefore, $m + n = 3 + 8 = 11$.

The final answer is $\boxed{11}$.