Solution Option 1. If the number $n$ does not end in 0, then the sum of the digits of $n-1$ is 1 less than the sum of the digits of $n$, which means it is 36. If the number $n$ ends in 0, then it ends in only one zero (since all digits in the decimal representation of $n$ are different). Then, if the number $n$ has the decimal representation $n=\overline{\ldots c b a 0}$, the number $n-1$ has the decimal representation $n-1=\overline{\ldots c b(a-1) 9}$, so the sum of the digits of $n-1$ is 8 more than that of $n$, which means it is 45.

To complete the solution, it remains to provide examples confirming that both answers are indeed possible. For the answer 36, the number 12345679 fits, and for the answer 45, the number 123456790 fits.
3.4. A natural number $n$ is written with different digits, the sum of which is 44. What can the sum of the digits of the number $n-1$ be? Find all possible options.

Answer, option 1. 36 and 45 (two answers).

Answer, option 2. 38 and 47 (two answers).

Answer, option 3. 41 and 50 (two answers).

Answer, option 4. 43 and 52 (two answers).