Find all numbers $n$ such, that in $[1;1000]$ there exists exactly $10$ numbers with digit sum equal to $n$.
To solve the problem, we need to find all numbers \( n \) such that there are exactly 10 numbers in the range \([1, 1000]\) whose digit sum equals \( n \).

1. **Understanding the digit sum**:
   - The digit sum of a number is the sum of its individual digits. For example, the digit sum of 123 is \(1 + 2 + 3 = 6\).

2. **Range of possible digit sums**:
   - The smallest number in the range \([1, 1000]\) is 1, with a digit sum of 1.
   - The largest number in the range is 999, with a digit sum of \(9 + 9 + 9 = 27\).
   - Therefore, the possible values of \( n \) range from 1 to 27.

3. **Counting numbers with a specific digit sum**:
   - We need to count how many numbers in the range \([1, 1000]\) have a digit sum equal to \( n \).

4. **Analyzing the given solution**:
   - The solution suggests considering numbers of the form \( abc \), \( aab \), and \( aaa \).
   - For \( abc \), there are \( 6 \) permutations (since \( a, b, c \) are distinct).
   - For \( aab \), there are \( 3 \) permutations (since two digits are the same).
   - For \( aaa \), there is \( 1 \) permutation (since all digits are the same).

5. **Checking specific values of \( n \)**:
   - For \( n = 3 \):
     - Possible combinations: \( 003, 012, 021, 030, 102, 111, 120, 201, 210, 300 \).
     - There are exactly 10 numbers with a digit sum of 3.
   - For \( n = 24 \):
     - Possible combinations: \( 789, 798, 879, 897, 978, 987 \).
     - There are exactly 6 numbers with a digit sum of 24.
   - For \( n = 27 \):
     - The only number is 999.
     - There is exactly 1 number with a digit sum of 27.

6. **Conclusion**:
   - The only values of \( n \) that result in exactly 10 numbers in the range \([1, 1000]\) with that digit sum are \( n = 3 \).

The final answer is \( \boxed{ n = 3 } \).