jmc

We know that $1.\overline{03}=1+0.\overline{03}$. Since $0.\overline{03}$ is three times greater than $0.\overline{01}$, we have $0.\overline{03}=3\cdot \frac{1}{99}=\frac{3}{99}$. This simplifies to $\frac{1}{33}$. Then, adding one to this fraction, we get $1+\frac{1}{33}=$ $\boxed{\frac{34}{33}}$.