$a_0, a_1, \ldots, a_{100}$ and $b_1, b_2,\ldots, b_{100}$ are sequences of real numbers, for which the property holds: for all $n=0, 1, \ldots, 99$, either
$$a_{n+1}=\frac{a_n}{2} \quad \text{and} \quad b_{n+1}=\frac{1}{2}-a_n,$$
or
$$a_{n+1}=2a_n^2 \quad \text{and} \quad b_{n+1}=a_n.$$
Given $a_{100}\leq a_0$, what is the maximal value of $b_1+b_2+\cdots+b_{100}$?
1. **Initial Assumptions and Setup:**
   We are given two sequences \(a_0, a_1, \ldots, a_{100}\) and \(b_1, b_2, \ldots, b_{100}\) with specific recursive properties. For each \(n = 0, 1, \ldots, 99\), either:
   \[
   a_{n+1} = \frac{a_n}{2} \quad \text{and} \quad b_{n+1} = \frac{1}{2} - a_n,
   \]
   or
   \[
   a_{n+1} = 2a_n^2 \quad \text{and} \quad b_{n+1} = a_n.
   \]
   We are also given that \(a_{100} \leq a_0\) and need to find the maximal value of \(b_1 + b_2 + \cdots + b_{100}\).

2. **Analyzing the Sequences:**
   - If \(a_0\) is negative, then \(a_{100}\) would be greater than \(a_0\), which contradicts \(a_{100} \leq a_0\). Therefore, \(a_0 \geq 0\).
   - Consider the case where \(a_{n+1} = 2a_n^2\) and \(b_{n+1} = a_n\). We use the inequality:
     \[
     a_n^2 \geq a_n - \frac{1}{4}
     \]
     which follows from the AM-GM inequality.

3. **Deriving the Inequality:**
   - Multiplying the inequality \(a_n^2 \geq a_n - \frac{1}{4}\) by 2, we get:
     \[
     2a_n^2 \geq 2a_n - \frac{1}{2}
     \]
   - This implies:
     \[
     a_{n+1} + \frac{1}{2} \geq 2a_n
     \]
   - Rearranging, we get:
     \[
     b_{n+1} \leq \frac{1}{2} + a_{n+1} - a_n
     \]

4. **Summing the Inequalities:**
   - Summing the inequalities for all \(n\) from 0 to 99, we get:
     \[
     \sum_{n=0}^{99} b_{n+1} \leq \sum_{n=0}^{99} \left( \frac{1}{2} + a_{n+1} - a_n \right)
     \]
   - This simplifies to:
     \[
     \sum_{n=0}^{99} b_{n+1} \leq 50 + a_{100} - a_0
     \]
   - Given \(a_{100} \leq a_0\), we have:
     \[
     \sum_{n=0}^{99} b_{n+1} \leq 50
     \]

5. **Example to Achieve the Bound:**
   - Consider \(a_i = \frac{1}{2}\) for every \(i\). Then:
     \[
     b_{n+1} = \frac{1}{2} - a_n = 0
     \]
   - This example shows that the sum can indeed reach 50.

Conclusion:
\[
\boxed{50}
\]