Find all three digit positive integers that have distinct digits and after their greatest digit is switched to $1$ become multiples of $30$.
1. Let \( N = \overline{abc} \) be a three-digit number where \( a, b, c \) are distinct digits.
2. We need to find \( N \) such that when its greatest digit is replaced by \( 1 \), the resulting number is a multiple of \( 30 \).

3. First, observe that \( N \) must end in \( 0 \) because a number that is a multiple of \( 30 \) must be a multiple of both \( 3 \) and \( 10 \). Therefore, \( c = 0 \).

4. Now, \( N \) can be written as \( \overline{ab0} \).

5. We need to consider two cases:
   - Case 1: \( b \) is the greatest digit.
   - Case 2: \( a \) is the greatest digit.

6. **Case 1: \( b \) is the greatest digit.**
   - After replacing \( b \) with \( 1 \), the number becomes \( \overline{a10} \).
   - For \( \overline{a10} \) to be a multiple of \( 30 \), it must be a multiple of \( 3 \) and \( 10 \). Since it ends in \( 0 \), it is already a multiple of \( 10 \).
   - To be a multiple of \( 3 \), the sum of its digits must be a multiple of \( 3 \). Therefore, \( a + 1 + 0 = a + 1 \) must be a multiple of \( 3 \).

7. **Case 2: \( a \) is the greatest digit.**
   - After replacing \( a \) with \( 1 \), the number becomes \( \overline{1b0} \).
   - For \( \overline{1b0} \) to be a multiple of \( 30 \), it must be a multiple of \( 3 \) and \( 10 \). Since it ends in \( 0 \), it is already a multiple of \( 10 \).
   - To be a multiple of \( 3 \), the sum of its digits must be a multiple of \( 3 \). Therefore, \( 1 + b + 0 = 1 + b \) must be a multiple of \( 3 \).

8. **Solving Case 1:**
   - \( a + 1 \) must be a multiple of \( 3 \).
   - Possible values for \( a \) are \( 2, 5, 8 \) (since \( a \) must be a digit and distinct from \( b \) and \( 0 \)).
   - Therefore, possible numbers are \( 230, 560, 890 \).

9. **Solving Case 2:**
   - \( 1 + b \) must be a multiple of \( 3 \).
   - Possible values for \( b \) are \( 2, 5, 8 \) (since \( b \) must be a digit and distinct from \( a \) and \( 0 \)).
   - Therefore, possible numbers are \( 320, 650, 980 \).

10. Combining both cases, the possible numbers are:
    - From Case 1: \( 230, 560, 890 \)
    - From Case 2: \( 320, 650, 980 \)

The final answer is \( \boxed{230, 560, 890, 320, 650, 980} \).