SAUDI ARABIAN IMO Booklet, 2023

Let $n = 2023$. First, we choose some $k$ such that $n \mid 10^k - 1$. For instance, any multiple of $\varphi(n)$ would work since $n$ is coprime to $10$. We still show that either $T = 10^k - 1$ or $T = 10^k - 2$ has the desired property, which completes the proof since $k$ can be taken to be arbitrarily large.

For this it suffices to show that $\#\{d_i(10^k - 1) : 1 \le i \le 9\} \le 2$. Indeed, if
$$\#\{d_i(10^k - 1) : 1 \le i \le 9\} = 1$$
then, since $10^k - 1$ which consists of all nines is a multiple of $n$, we have
$$d_i(10^k - 2) = d_i(10^k - 1) \text{ for } i \in \{1, \dots, 8\}, \text{ and } d_9(10^k - 2) < d_9(10^k - 1).$$
This means that $\#\{d_i(10^k - 2) : 1 \le i \le 9\} = 2$.

To prove that $\#\{d_i(10^k - 1)\} \le 2$ we need an observation. Now let
$$\overline{a_{k-1}a_{k-2}\dots a_0} \in \{1, \dots, 10^k - 1\}$$
be the decimal expansion of an arbitrary number, possibly with leading zeroes. Then $\overline{a_{k-1}a_{k-2}\dots a_0}$ is divisible by $n$ if and only if $\overline{a_{k-2}\dots a_0a_{k-1}}$ is divisible by $n$. Indeed, this follows from the fact that
$$10 \cdot \overline{a_{k-1}a_{k-2}\dots a_0} - \overline{a_{k-2}\dots a_0a_{k-1}} = (10^k - 1) \cdot a_{k-1}$$
is divisible by $n$. This observation shows that the set of multiples of $n$ between $1$ and $10^k - 1$ is invariant under simultaneous cyclic permutation of digits when numbers are written with leading zeroes.

Hence, for each $i \in \{1, \dots, 9\}$ the number $d_i(10^k - 1)$ is $k$ times larger than the number of $k$-digit numbers which start from the digit $i$ and are divisible by $n$. Since the latter number is either $\lfloor \frac{10^{k-1}}{n} \rfloor$ or $1 + \lfloor \frac{10^{k-1}}{n} \rfloor$, we conclude that $\#\{d_i(10^k - 1)\} \le 2$. This finishes the solution. $\square$