jmc

Converting from a decimal to a fraction, we obtain $0.abc=\frac{abc}{1000}=\frac{abc}{2^3\cdot 5^3}=\frac{1}{y}$. Since $0<y\le 9$ and $y$ divides into $1000$, $y$ must equal one of $1,2,4,5$ or $8$. Notice that $y\ne 1$ because then $abc=1000$, which is impossible as $a$, $b$, and $c$ are digits. So, breaking down the remaining possibilities: $$\begin{array}{rl}y& =2\Rightarrow abc=2^2\cdot 5^3=500\Rightarrow a+b+c=5+0+0=5\\ y& =4\Rightarrow abc=2\cdot 5^3=250\Rightarrow a+b+c=2+5+0=7\\ y& =5\Rightarrow abc=2^3\cdot 5^2=200\Rightarrow a+b+c=2+0+0=2\\ y& =8\Rightarrow abc=5^3=125\Rightarrow a+b+c=1+2+5=8.\end{array}$$Therefore, the largest possible value of $a+b+c$ is $\boxed{8}$.