How many ways are there to permute the first $n$ positive integers such that in the permutation, for each value of $k \le n$, the first $k$ elements of the permutation have distinct remainder mod $k$?
To solve the problem, we need to find the number of permutations of the first \( n \) positive integers such that for each \( k \le n \), the first \( k \) elements of the permutation have distinct remainders when taken modulo \( k \).

1. **Define the problem in terms of \( P(n) \):**
   Let \( P(n) \) denote the number of such permutations of the first \( n \) positive integers.

2. **Base cases:**
   - For \( n = 1 \), there is only one permutation: \((1)\). Thus, \( P(1) = 1 \).
   - For \( n = 2 \), the permutations are \((1, 2)\) and \((2, 1)\). Both satisfy the condition. Thus, \( P(2) = 2 \).

3. **Recursive step:**
   Consider a permutation \((x_1, x_2, \ldots, x_n)\) that satisfies the condition. We need to determine the possible values for \( x_n \):
   - If \( x_n = n \), then the first \( n-1 \) elements must form a valid permutation of \(\{1, 2, \ldots, n-1\}\). There are \( P(n-1) \) such permutations.
   - If \( x_n = 1 \), then the first \( n-1 \) elements must form a valid permutation of \(\{2, 3, \ldots, n\}\) with the additional condition that the first \( n-1 \) elements have distinct remainders modulo \( n-1 \). This is equivalent to finding a valid permutation of \(\{2, 3, \ldots, n\}\) and then appending \( 1 \) at the end. There are \( P(n-1) \) such permutations.

4. **Considering the cases for \( x_{n-1} \):**
   - If \( x_{n-1} = n \), then the remaining elements \(\{2, 3, \ldots, n-1\}\) must form a valid permutation. There are \( P(n-2) \) such permutations.
   - If \( x_{n-1} = 2 \), then the remaining elements \(\{3, 4, \ldots, n-1\}\) must form a valid permutation. There are \( P(n-2) \) such permutations.

5. **Combining the cases:**
   - If \( x_n = n \), there are \( P(n-1) \) permutations.
   - If \( x_n = 1 \), there are \( 2P(n-2) \) permutations (one for \( x_{n-1} = n \) and one for \( x_{n-1} = 2 \)).

6. **Recursive formula:**
   Combining these, we get:
   \[
   P(n) = P(n-1) + 2P(n-2)
   \]

7. **Solving the recurrence relation:**
   We solve the recurrence relation \( P(n) = P(n-1) + 2P(n-2) \) with initial conditions \( P(1) = 1 \) and \( P(2) = 2 \).

   By solving this recurrence relation, we find that:
   \[
   P(n) = 2^{n-1}
   \]

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