We claim the answer is no. Let $S$ be the set of positive integers without three $1$s in their decimal representation. If there existed such a sequence, we would have to have $\frac{|S\cap\{1,...,10^d-1\}|}{10^d-1}>\frac{1}{1999}$ for all positive integers $d$, but we will prove that the LHS approaches $0$ as $d$ goes to infinity. It's not hard to compute that $|S\cap\{1,...,10^d-1\}|=\binom{d}{2}9^{d-2}+\binom{d}{1}9^{d-1}+9^d-1$. Therefore, $$\lim_{d\rightarrow +\infty}\frac{|S\cap\{1,...,10^d-1\}|}{10^d-1}=\lim_{d\rightarrow +\infty}\frac{\binom{d}{2}9^{d-2}+\binom{d}{1}9^{d-1}+9^d-1}{10^d-1}=0$$(notice that $\binom{d}{2}$ and $\binom{d}{1}$ are polynomials which grow much slower than exponentials). Therefore we have a contradiction and no such sequence exists.