Let $k = 2021$. We want to construct infinitely many positive integers divisible by $k$ such that in their decimal representation, each digit $0, 1, \ldots, 9$ appears the same number of times.

Let $n$ be a positive integer. Consider the number $N$ whose decimal representation consists of $n$ copies of each digit $0, 1, \ldots, 9$ (in any order, but for definiteness, let us take the number $M_n$ formed by writing $0, 1, 2, \ldots, 9$ in order, $n$ times, i.e., $01234567890123456789\ldots$ repeated $n$ times, for a total of $10n$ digits).

Let $S$ be the set of all such numbers $M_n$ for $n \geq 1$ (note that $M_n$ may have leading zeros, but we can permute the digits to avoid this, or simply consider all numbers with $n$ copies of each digit).

There are $\binom{10n}{n, n, \ldots, n}$ such numbers (the multinomial coefficient), so for each $n$, the set $S_n$ of numbers with $n$ copies of each digit is finite but very large.

Now, $k = 2021$ is fixed. For each $n$, consider the set $S_n$ modulo $k$. Since $|S_n|$ grows rapidly with $n$, and there are only $k$ possible residues modulo $k$, by the pigeonhole principle, for sufficiently large $n$, there must exist at least one number in $S_n$ divisible by $k$.

More precisely, for each $n$, there exists at least one number with $n$ copies of each digit that is divisible by $k$. Since $n$ can be taken arbitrarily large, there are infinitely many such numbers.

Therefore, there exist infinitely many positive integers divisible by $2021$ and each of them contains the same number of digits $0, 1, \ldots, 9$.