Answer: 669

Small Hint:

Expand each $\frac{1}{10^k - 1}$ as a geometric series in $10^{-k}$ and collect equal powers of $10:$ the coefficient of $10^{-n}$ counts the divisors of $n.$

Big Hint:

After multiplying by $10^{100},$ only $n = 98, 99, 100$ affect the last three digits, and the leftover tail is between $0$ and $1.$

Solution:

Each term is $\frac{1}{10^k - 1} = \sum_{j \ge 1} 10^{-kj},$ so summing over $k$ and collecting the exponent $n = kj,$ $$S = \sum_{n \ge 1} \frac{d(n)}{10^n},$$ where $d(n)$ is the number of divisors of $n.$ Hence $10^{100} S = \sum_{n = 1}^{100} d(n)\,10^{100 - n}$ $+ T$ with $T = \sum_{m \ge 1} d(100 + m)\,10^{-m}.$

From $d(101) = 2,$ $d(102) = 8,$ $d(103) = 2,$ $d(104) = 8,$ the tail starts $0.2$ $+ 0.08$ $+ 0.002$ $+ 0.0008 = 0.2828,$ and since $d(N) \lt 2\sqrt{N},$ the remaining terms contribute less than $\sum_{m \ge 5} \frac{2\sqrt{100 + m}}{10^m} \lt 0.001.$ So $0 \lt T \lt 1$ and $$\left\lfloor 10^{100} S \right\rfloor = \sum_{n = 1}^{100} d(n)\,10^{100 - n}.$$

Modulo $1000,$ every term with $n \le 97$ is a multiple of $1000,$ leaving $d(98) \cdot 100 + d(99) \cdot 10$ $+ d(100).$ Since $d(98) = 6,$ $d(99) = 6,$ and $d(100) = 9,$ the remainder is $600 + 60 + 9 = 669.$