We know that there exists a positive integer with $7$ distinct digits which is multiple of each of them. What are its digits?

(Paolo Leonetti)
1. Let \( n \) be the positive integer with 7 distinct digits, and let \( S \) be the set of its digits. Since \( n \) is a positive integer, \( 0 \notin S \) because no positive integer can be a multiple of \( 0 \).

2. Note that \( S \) must contain an even number. If \( S \) did not contain an even number, it would have at most five elements (1, 3, 5, 7, 9), which is less than 7. Therefore, \( S \) must contain at least one even number.

3. Since \( S \) must contain an even number, \( 5 \notin S \). If \( 5 \) were in \( S \), \( n \) would be a multiple of both \( 5 \) and some even number, which would make \( n \) end in \( 0 \). This is not possible because \( 0 \notin S \).

4. Consider the sum of the digits in \( S \). If \( 9 \) were not in \( S \), the sum of the digits would be:
   \[
   1 + 2 + 3 + 4 + 6 + 7 + 8 = 31
   \]
   Since 31 is not a multiple of 9, \( n \) would not be divisible by 9, which contradicts the requirement that \( n \) is a multiple of each of its digits. Therefore, \( 9 \in S \).

5. Since \( 9 \in S \), the sum of the digits in \( S \) must be a multiple of 9. The sum of all digits from 1 to 9 is:
   \[
   1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45
   \]
   To make the sum of the digits in \( S \) a multiple of 9, we need to exclude one digit such that the remaining sum is still a multiple of 9. We need to exclude a digit such that the sum of the remaining digits is 36 (since 45 - 36 = 9, which is a multiple of 9).

6. Excluding 4 from the set of digits gives:
   \[
   1 + 2 + 3 + 6 + 7 + 8 + 9 = 36
   \]
   Since 36 is a multiple of 9, the set \( S \) must be:
   \[
   S = \{1, 2, 3, 6, 7, 8, 9\}
   \]

The final answer is \( \boxed{ \{1, 2, 3, 6, 7, 8, 9\} } \)