The answer is $50$. If $a_0 < 0$ then it is clear that $a_{100} > a_0$. If $a_0 = 0$, then $b_n \in \{0, 1/2\}$ and hence $S = b_1 + b_2 + \dots + b_{100} \le 50$. Assume that $a_0 > 0$. In this case all terms of the sequence $(a_n)$ are positive. We have
$$(a_n - \frac{a_{n-1}}{2})(a_n - 2a_{n-1}^2) = 0 \quad \text{and} \quad a_{100} \le a_0.$$
This equation can be expressed as
$$\frac{a_n}{a_{n-1}} + \frac{a_{n-1}^2}{a_n} = 2a_{n-1} + \frac{1}{2}.$$

Summing up the equations for $n = 1, 2, ..., 100$ side by side, we get
$$\sum_{n=1}^{100} \frac{a_n}{a_{n-1}} + \sum_{n=1}^{100} \frac{a_{n-1}^2}{a_n} = 2 \sum_{n=1}^{100} a_{n-1} + 50.$$
Using Cauchy-Schwarz Inequality
$$\sum_{n=1}^{100} \frac{a_{n-1}^2}{a_n} \ge \frac{(a_0 + a_1 + \dots + a_{99})^2}{a_1 + a_2 + \dots + a_{100}} \ge a_0 + a_2 + \dots + a_{99} = \sum_{n=1}^{100} a_{n-1}.$$
This yields that
$$\sum_{n=1}^{100} \left( \frac{a_n}{a_{n-1}} - a_{n-1} \right) \le 50.$$
The rule given in the problem implies that $b_n = \frac{a_n}{a_{n-1}} - a_{n-1}$ for all $n = 1, 2, ..., 100$. Hence $S \le 50$.
The equality holds when $a_i = 1/2, i = 0, 1, ..., 100, b_i = 1/2, i = 1, 2, ..., 100$.