Determine all the positive integers with more than one digit, all distinct, such that the sum of its digits is equal to the product of its digits.

1. **Identify the problem constraints:**
   - We need to find positive integers with more than one digit.
   - All digits must be distinct.
   - The sum of the digits must equal the product of the digits.

2. **Consider 2-digit numbers:**
   - Let the number be \( \overline{ab} \) where \( a \) and \( b \) are distinct digits.
   - The sum of the digits is \( a + b \).
   - The product of the digits is \( ab \).
   - We need \( a + b = ab \).

   Rearrange the equation:
   \[
   ab - a - b = 0 \implies ab - a - b + 1 = 1 \implies (a-1)(b-1) = 1
   \]
   The only integer solutions for \( (a-1)(b-1) = 1 \) are \( (a-1, b-1) = (1, 1) \), which implies \( a = 2 \) and \( b = 2 \). However, \( a \) and \( b \) must be distinct, so there are no 2-digit solutions.

3. **Consider 3-digit numbers:**
   - Let the number be \( \overline{abc} \) where \( a, b, \) and \( c \) are distinct digits.
   - The sum of the digits is \( a + b + c \).
   - The product of the digits is \( abc \).
   - We need \( a + b + c = abc \).

   Check possible combinations:
   - For \( a = 1, b = 2, c = 3 \):
     \[
     1 + 2 + 3 = 6 \quad \text{and} \quad 1 \cdot 2 \cdot 3 = 6
     \]
     This is a valid solution.

   - Permutations of \( 1, 2, 3 \) are also valid:
     \[
     123, 132, 213, 231, 312, 321
     \]

4. **Consider 4-digit numbers:**
   - Let the number be \( \overline{abcd} \) where \( a, b, c, \) and \( d \) are distinct digits.
   - The sum of the digits is \( a + b + c + d \).
   - The product of the digits is \( abcd \).
   - We need \( a + b + c + d = abcd \).

   By intense casework, it can be shown that no such combinations exist.

5. **Consider 5 or more digit numbers:**
   - For \( n \)-digit numbers, the product of the digits must be at least \( n! \) (factorial of \( n \)).
   - The sum of the digits must be less than or equal to \( 9n \) (since the maximum digit is 9).
   - For \( n \geq 5 \):
     \[
     n! > 9n
     \]
     This inequality holds true for all \( n \geq 5 \), thus no such numbers exist.

Conclusion:
The only valid solutions are the permutations of 123.

The final answer is \( \boxed{ \{123, 132, 213, 231, 312, 321\} } \)