No, it is not possible. Indeed, let $n_i$ be the number of 01-sequences in the set with 1 at the $i$th entry, where $1 \le i \le 200$. We also let $S$ be the total number of different positions between all pairs of 01-sequences in the set. Since every pair differs at least at 101 positions, we have $S \ge 101 \binom{n}{2}$. On the other hand, there are $n_i(n - n_i)$ pairs of different digits in the $i$th entry. Thus, by considering the $i$th positions of all pairs of 01-sequences, we have $S = \sum_{i=1}^{200} n_i(n - n_i)$. Using the AM-GM inequality, we have
$$n_i(n - n_i) \le \left( \frac{n_i + (n - n_i)}{2} \right)^2 = \frac{n^2}{4}.$$
Hence we have
$$50n^2 = \sum_{i=1}^{200} \frac{n^2}{4} \ge S \ge 101 \binom{n}{2}.$$
This implies $n \le 101$. However, when $n = 101$, the inequality $\frac{n^2}{4} \ge n_i(n - n_i)$ is strict, as one side is an integer and the other side is not. Thus $n < 101$, or $n \le 100$.