The cards in a stack of $2n$ cards are numbered consecutively from $1$ through $2n$ from top to bottom. The top $n$ cards are removed, kept in order, and form pile $A$. The remaining cards form pile $B$. The cards are then restacked by taking cards alternately from the tops of pile $B$ and $A$, respectively. In this process, card number $(n+1)$ becomes the bottom card of the new stack, card number $1$ is on top of this card, and so on, until piles $A$ and $B$ are exhausted. If, after the restacking process, at least one card from each pile occupies the same position that it occupied in the original stack, the stack is named [i]magical[/i]. Find the number of cards in the magical stack in which card number $131$ retains its original position.
1. **Understanding the Problem:**
   - We have a stack of \(2n\) cards numbered from 1 to \(2n\).
   - The top \(n\) cards are removed to form pile \(A\), and the remaining \(n\) cards form pile \(B\).
   - The cards are then restacked by taking cards alternately from the tops of pile \(B\) and \(A\).
   - We need to find the number of cards in the stack such that card number 131 retains its original position after restacking.

2. **Restacking Process:**
   - Let's denote the original positions of the cards as \(1, 2, 3, \ldots, 2n\).
   - After restacking, the new positions of the cards can be represented as follows:
     - The first card from pile \(B\) (originally at position \(n+1\)) goes to the bottom.
     - The first card from pile \(A\) (originally at position 1) goes on top of it.
     - The second card from pile \(B\) (originally at position \(n+2\)) goes next, and so on.

3. **Creating the Array:**
   - We create an array to represent the original and new positions of the cards:
     \[
     \begin{array}{c|c}
     \text{Original} & \text{New} \\
     1 & n \\
     2 & 2n \\
     3 & n-1 \\
     4 & 2n-1 \\
     5 & n-2 \\
     6 & 2n-2 \\
     \vdots & \vdots \\
     2n-1 & 1 \\
     2n & n+1 \\
     \end{array}
     \]

4. **Case Analysis:**
   - **Case 1:** Card 131 comes from the top \(n\) cards (pile \(A\)).
     - The new position of card \(k\) from pile \(A\) is \(2(n-k+1)\).
     - We need to find \(k\) such that \(2(n-k+1) = 131\).
     - Solving for \(k\):
       \[
       2(n-k+1) = 131 \implies n-k+1 = \frac{131}{2} \implies n-k+1 = 65.5
       \]
       - Since \(n-k+1\) must be an integer, this case is not possible.

   - **Case 2:** Card 131 comes from the bottom \(n\) cards (pile \(B\)).
     - The new position of card \(k\) from pile \(B\) is \(2k-1\).
     - We need to find \(k\) such that \(2k-1 = 131\).
     - Solving for \(k\):
       \[
       2k-1 = 131 \implies 2k = 132 \implies k = 66
       \]
       - Therefore, card 131 is originally at position \(n+66\).

5. **Finding \(n\):**
   - Since card 131 is originally at position \(n+66\), we have:
     \[
     n+66 = 131 \implies n = 131 - 66 \implies n = 65
     \]
   - The total number of cards in the stack is \(2n = 2 \times 65 = 130\).

The final answer is \(\boxed{130}\).