Determine all positive integers $d$ such that whenever $d$ divides a positive integer $n$, $d$ will also divide any integer obtained by rearranging the digits of $n$.
To determine all positive integers \( d \) such that whenever \( d \) divides a positive integer \( n \), \( d \) will also divide any integer obtained by rearranging the digits of \( n \), we can proceed as follows:

1. **Consider the property of divisibility by \( d \):**
   Let \( n \) be a positive integer with digits \( a_1, a_2, \ldots, a_k \). We can express \( n \) in its decimal form as:
   \[
   n = a_1 \cdot 10^{k-1} + a_2 \cdot 10^{k-2} + \cdots + a_k
   \]
   If \( d \) divides \( n \), then \( d \) must also divide any number obtained by rearranging the digits of \( n \).

2. **Rearranging the digits:**
   Consider a rearrangement of the digits of \( n \), say \( m \). Then \( m \) can be written as:
   \[
   m = b_1 \cdot 10^{k-1} + b_2 \cdot 10^{k-2} + \cdots + b_k
   \]
   where \( \{b_1, b_2, \ldots, b_k\} \) is a permutation of \( \{a_1, a_2, \ldots, a_k\} \).

3. **Difference between \( n \) and \( m \):**
   Consider the difference \( n - m \):
   \[
   n - m = (a_1 \cdot 10^{k-1} + a_2 \cdot 10^{k-2} + \cdots + a_k) - (b_1 \cdot 10^{k-1} + b_2 \cdot 10^{k-2} + \cdots + b_k)
   \]
   Since \( \{b_1, b_2, \ldots, b_k\} \) is a permutation of \( \{a_1, a_2, \ldots, a_k\} \), the sum of the digits remains the same, but the positional values change.

4. **Divisibility by 9:**
   Notice that the sum of the digits of \( n \) and \( m \) is the same. Therefore, \( n \equiv m \pmod{9} \). This implies that \( n - m \) is divisible by 9:
   \[
   n - m \equiv 0 \pmod{9}
   \]
   Hence, \( d \) must divide 9.

5. **Possible values of \( d \):**
   The divisors of 9 are \( 1, 3, \) and \( 9 \). We need to check if these values satisfy the condition:
   - For \( d = 1 \): Any integer divides any rearrangement of its digits.
   - For \( d = 3 \): Any integer divisible by 3 will have a sum of digits divisible by 3, and any rearrangement will also be divisible by 3.
   - For \( d = 9 \): Any integer divisible by 9 will have a sum of digits divisible by 9, and any rearrangement will also be divisible by 9.

Thus, the values of \( d \) that satisfy the condition are \( 1, 3, \) and \( 9 \).

The final answer is \( \boxed{d \in \{1, 3, 9\}} \)