Vladimir writes the numbers from 1 to 1000 on the board. As long as there is a number strictly greater than 9 written on the board, he chooses whichever he wants, and replaces it with the sum of its digits. How many times can he end up with the number 1?
Let's already note that Vladimir will eventually end up with only numbers between 0 and 9, since each time, the number is replaced by a smaller number: indeed, any number with at least two digits is strictly greater than the sum of its digits.

Indeed, if $n=\overline{a_{k} a_{k-1} \ldots a_{0}}$ with $k \geqslant 1$ and $a_{k} \neq 0, n=10^{k} a_{k}+\cdots+10^{1} a_{1}+10^{0} a_{0}>a_{k}+\cdots+a_{0}$ (the inequality being strict because $10^{k} a_{k}>a_{k}$).

Let's recall, as seen in the modulo course, that a number is congruent to the sum of its digits modulo 9. In particular, at each operation, the number of elements in the array that are congruent to 1 modulo 9 is invariant. At the end, every element written being between 0 and 9, the number of elements equal to 1 is the number of elements congruent to 1 modulo 9. It remains to calculate how many elements between 1 and 1000 are congruent to 1 modulo $n$, i.e., can be written in the form $9 k+1$ with $k$ an integer.

The inequality $1 \leqslant 9 k+1 \leqslant 1000$ is equivalent to $0 \leqslant 9 k \leqslant 999$, i.e., to $0 \leqslant k \leqslant 111$, so there are 112 possible values of $k$, and since two different values of $k$ give two different values of $9 k+1$ between 1 and 1000. Therefore, there are initially 112 numbers congruent to 1 modulo 9, so at the end, there are 112 numbers equal to 1.