Let $n$ be a positive integer. Let $B_n$ be the set of all binary strings of length $n$. For a binary string $s_1\hdots s_n$, we define it's twist in the following way. First, we count how many blocks of consecutive digits it has. Denote this number by $b$. Then, we replace $s_b$ with $1-s_b$. A string $a$ is said to be a [i]descendant[/i] of $b$ if $a$ can be obtained from $b$ through a finite number of twists. A subset of $B_n$ is called [i]divided[/i] if no two of its members have a common descendant. Find the largest possible cardinality of a divided subset of $B_n$.

[i]Remark.[/i] Here is an example of a twist: $101100 \rightarrow 101000$ because $1\mid 0\mid 11\mid 00$ has $4$ blocks of consecutive digits. 

[i]Viktor Simjanoski[/i]
1. **Define the Problem and Initial Setup:**
   - Let \( n \) be a positive integer.
   - Let \( B_n \) be the set of all binary strings of length \( n \).
   - For a binary string \( s_1s_2\ldots s_n \), define its twist by counting the number of blocks of consecutive digits, denoted by \( b \). Replace \( s_b \) with \( 1 - s_b \).
   - A string \( a \) is a descendant of \( b \) if \( a \) can be obtained from \( b \) through a finite number of twists.
   - A subset of \( B_n \) is divided if no two of its members have a common descendant.
   - We aim to find the largest possible cardinality of a divided subset of \( B_n \).

2. **Graph Representation:**
   - Consider a directed graph where nodes are binary strings of length \( n \) and directed edges represent the twist operation.
   - Each node has exactly one outgoing edge, forming a directed graph where each connected component contains exactly one simple cycle.

3. **Cycle Length in Connected Components:**
   - Claim: Every connected component has a cycle of length 2.
   - **Proof:**
     - Consider how the number of blocks changes with a twist. If the string has more than 1 but less than \( n \) blocks, the number of blocks can change by \( \pm 2 \) or remain the same.
     - For strings like \( 00\ldots0 \) and \( 11\ldots1 \), the cycle is \( 00\ldots0 \rightarrow 10\ldots0 \rightarrow 00\ldots0 \), forming a cycle of length 2.
     - For alternating strings like \( 1010\ldots \) and \( 0101\ldots \), the cycle is \( 1010\ldots \rightarrow 1000\ldots \rightarrow 1010\ldots \), again forming a cycle of length 2.
     - For other strings, assume a cycle length greater than 2. If adjacent nodes have the same number of blocks, it forms a cycle of length 2, contradicting the assumption.
     - If adjacent nodes have different block counts, the number of nodes in the cycle is even. Let \( t \) be the maximum number of blocks. The element at position \( t \) changes an even number of times, leading to a contradiction if \( t \) is maximal.

4. **Counting Connected Components:**
   - Each connected component corresponds to a cycle of length 2.
   - To count the number of such cycles, consider strings with \( i \) blocks where a twist results in \( i \) blocks again.
   - This occurs if \( s_{i-1} = s_i \neq s_{i+1} \) or \( s_{i-1} \neq s_i = s_{i+1} \).
   - The total number of such strings is \( 2^{n-2} \).

5. **Conclusion:**
   - The largest possible cardinality of a divided subset of \( B_n \) is \( 2^{n-2} \).

The final answer is \( \boxed{2^{n-2}} \).