Let $n$ be a positive integer. Initially, a bishop is placed in each square of the top row of a $2^n \times 2^n$
chessboard; those bishops are numbered from $1$ to $2^n$ from left to right. A [i]jump[/i] is a simultaneous move made by all bishops such that each bishop moves diagonally, in a straight line, some number of squares, and at the end of the jump, the bishops all stand in different squares of the same row.

Find the total number of permutations $\sigma$ of the numbers $1, 2, \ldots, 2^n$ with the following property: There exists a sequence of jumps such that all bishops end up on the bottom row arranged in the order $\sigma(1), \sigma(2), \ldots, \sigma(2^n)$, from left to right.

[i]Israel[/i]
1. **Understanding the Problem:**
   We need to find the number of permutations $\sigma$ of the numbers $1, 2, \ldots, 2^n$ such that there exists a sequence of jumps that moves the bishops from the top row to the bottom row of a $2^n \times 2^n$ chessboard, with the bishops ending up in the order $\sigma(1), \sigma(2), \ldots, $\sigma(2^n)$.

2. **Characterizing the Jumps:**
   Each jump must move the bishops diagonally such that they end up in different squares of the same row. Let’s denote the length of a jump as the number of squares moved vertically. We claim that the length of a jump must be $2^s$ for some $0 \leq s \leq n-1$.

3. **Proof of Jump Length:**
   Suppose a jump has length $r$. Then, we have a permutation $\pi: \{1, 2, \cdots, 2^n\} \mapsto \{1, 2, \cdots, 2^n\}$ such that $\pi(x) \in \{x-r, x+r\}$ for every $x \in \{1, 2, \cdots, 2^n\}$. Since $\pi$ must be a bijection, the set $\{1, 2, \cdots, 2^n\}$ can be split into intervals of length $2r$, implying $2r \mid 2^n$. Therefore, $r \mid 2^{n-1}$, which means $r$ must be of the form $2^s$ for some $0 \leq s \leq n-1$.

4. **Involutions and Commutativity:**
   Let the permutation corresponding to a jump of length $2^s$ be $\pi_s$. Each $\pi_s$ is an involution, meaning $\pi_s^2 = \text{id}$. We also claim that $\pi_r(\pi_s(x)) = \pi_s(\pi_r(x))$ for all $x \in \{1, 2, \cdots, 2^n\}$, meaning the permutations commute.

5. **Proof of Commutativity:**
   We consider four cases based on the values of $\pi_s(x)$ and $\pi_r(x)$:
   - If $\pi_s(x) = x + 2^s$ and $\pi_r(x) = x + 2^r$, then $\pi_r(\pi_s(x)) = x + 2^s + 2^r$ and $\pi_s(\pi_r(x)) = x + 2^r + 2^s$.
   - If $\pi_s(x) = x - 2^s$ and $\pi_r(x) = x + 2^r$, then $\pi_r(\pi_s(x)) = x - 2^s + 2^r$ and $\pi_s(\pi_r(x)) = x + 2^r - 2^s$.
   - If $\pi_s(x) = x + 2^s$ and $\pi_r(x) = x - 2^r$, then $\pi_r(\pi_s(x)) = x + 2^s - 2^r$ and $\pi_s(\pi_r(x)) = x - 2^r + 2^s$.
   - If $\pi_s(x) = x - 2^s$ and $\pi_r(x) = x - 2^r$, then $\pi_r(\pi_s(x)) = x - 2^s - 2^r$ and $\pi_s(\pi_r(x)) = x - 2^r - 2^s$.

   In all cases, $\pi_r(\pi_s(x)) = \pi_s(\pi_r(x))$, proving the commutativity.

6. **Counting the Permutations:**
   Each permutation $\sigma$ can be written as a composition of $\pi_s$ for $0 \leq s \leq n-1$. Since each $\pi_s$ is an involution and the order of composition does not matter, we can choose each $\pi_s$ to appear either 0 or 1 times. This gives us $2^{n-1}$ possible permutations.

7. **Verification:**
   To verify that all these permutations can be achieved, note that the sum of the lengths of the jumps must equal $2^n - 1$. This can be achieved by a combination of jumps of lengths $2^s$ for $0 \leq s \leq n-1$.

Thus, we have shown that there are exactly $2^{n-1}$ permutations $\sigma$.

The final answer is $\boxed{2^{n-1}}$.