Croatian Mathematical Competitions

Let $n = \overline{a_{m-1}a_{m-2}\dots a_1a_0}$. We know that $a_0 > a_1 > \dots > a_{m-1}$.

Since $9n = 10n - n$, from
$$\text{a m-1 a m-2 a m-3 a 1 a 0 0 - a m-1 a m-2 a 2 a 1 a 0 b m b m-1 b m-2 b 2 b 1 b 0}$$
we conclude $b_0 = 10 - a_0$, $b_1 = a_0 - (a_1 + 1)$, $b_2 = a_1 - a_2$, $b_3 = a_2 - a_3$, ..., $b_{m-1} = a_{m-2} - a_{m-1}$, $b_m = a_{m-1}$. Those are the digits of $9n$, and their sum is $a_{m-1} + (a_{m-2} - a_{m-1}) + \dots + (a_1 - a_2) + (a_0 - a_1 - 1) + (10 - a_0) = 9$.