smc

$$\begin{gathered} \frac{1}{99^2} \\ =\frac{1}{99}\cdot \frac{1}{99} \\ =\frac{0.\overline{01}}{99} \\ =0.\overline{\mathrm{00010203...9799}} \end{gathered}$$ So, the answer is $0+0+0+1+0+2+0+3+...+9+7+9+9=2\cdot 10\cdot \frac{9\cdot 10}{2}-(9+8)$ or $\boxed{883}$. There are two things to notice here. First, $\frac{1}{99}$ has a very simple and unique decimal expansion, as shown. Second, for $\frac{0.\overline{01}}{99}$ to itself produce a repeating decimal, $99$ has to evenly divide a sufficiently extended number of the form $\mathrm{101010101..}$. This number will have $99$ ones (197 digits in total), as to be divisible by $9$ and $11$. The enormity of this number forces us to look for a pattern, and so we divide out as shown. Indeed, upon division (seeing how the remainders always end in "501" or "601" or, at last, "9801"), we find the repeating part $.00010203040506070809101112131415....9799$. If we wanted to further check our pattern, we could count the total number of digits in our quotient (not counting the first three): 195. Since $99<100$, multiplying by it will produce either $1$ or $2$ extra digits, so our quotient passes the test. Or note that the 1 has to be carried when you get to 100 so the 99 becomes 00 and the 1 gets carried again to 98 which becomes 99.