jmc

We write each repeating decimal as a fraction. We convert $0.\overline{1}$ to a fraction by setting $0.\overline{1}=x$. Multiplying both sides by 10, we get $10x=1.\overline{1}$. We subtract those two equations to get $$\begin{array}{rl}10x-x& =1.\overline{1}-0.\overline{1}⟹\\ 9x& =1⟹\\ x& =\frac{1}{9}.\end{array}$$Next, we convert $0.\overline{02}$ to a fraction by setting $0.\overline{02}=y$. Multiplying by 100, we get $100y=2.\overline{02}$. We subtract those two equations to get: $$\begin{array}{rl}100y-y& =2.\overline{02}-0.\overline{02}⟹\\ 99y& =2⟹\\ y& =\frac{2}{99}.\end{array}$$Finally, we convert $0.\overline{003}$ to a fraction by setting $0.\overline{003}=z$. Multiplying by 1000, we get $1000z=3.\overline{003}$. We subtract those two numbers to get: $$\begin{array}{rl}1000z-z& =3.\overline{003}-0.\overline{003}⟹\\ 999z& =3⟹\\ z& =\frac{3}{999}.\end{array}$$The requested sum is $\frac{1}{9}+\frac{2}{99}+\frac{3}{999}=\boxed{\frac{164}{1221}}$.