smc

Let us begin by working with the condition $0.\overline{ab}=0.ababab\cdots ,$. Let $x=0.ababab\cdots$. So, $100x-x=ab\Rightarrow x=\frac{ab}{99}$. In order for this fraction $x$ to be in the form $\frac{1}{n}$, $99$ must be a multiple of $ab$. Hence the possibilities of $ab$ are $1,3,9,11,33,99$. Checking each of these, $\frac{1}{99}=0.\overline{01},\frac{3}{99}=\frac{1}{33}=0.\overline{03},\frac{9}{99}=\frac{1}{11}=0.\overline{09},\frac{11}{99}=\frac{1}{9}=0.\overline{1},\frac{33}{99}=\frac{1}{3}=0.\overline{3},$ and $\frac{99}{99}=1$. So the only values of $n$ that have distinct $a$ and $b$ are $11,33,$ and $99$. So, $11+33+99=\boxed{143}$