We denote $c_i = a_i - i$, then $0 \le c_1 \le c_2 \le ... \le c_n \le 100 - n$, $b_i = c_i$ or $b_i = c_i + n$ for all $i = 1, n$. Then, $c_i + n \le c_n + n = a_n \le 100$. Note that among the three consecutive numbers $c_i, c_{i+1}$ and $c_{i+2}$, at least two are different, because otherwise, at least two out of three numbers $b_i, b_{i+1}$ and $b_{i+2}$ would be the same, which contradicts problem statement. Hence, $c_{i+2} \ge c_i + 1$.

Thus, $c_n \ge c_{n-2} + 1 \ge \dots \ge \frac{1}{2}(n-1)$. On the other hand, $c_n \le 100 - n$. Thus, $\frac{1}{2}(n-1) \le c_n \le 100 - n \Rightarrow n \le 67$. Now we show an example for $n=67$:

$$c_1 = c_2 = 0,\ c_3 = c_4 = 1,\ \dots,\ c_{65} = c_{66} = 32\ \text{ and }\ c_{67} = 33,$$
$$b_1 = 0,\ b_2 = 67,\ b_3 = 1,\ b_4 = 68,\ \dots,\ b_{65} = 32,\ b_{66} = 99\ \text{ and }\ b_{67} = 33,$$
therefore, $a_i = c_i + i$.