68. National Mathematical Olympiad

Answer: $1\le a<2019$.

It is clear that $a\ge 1$, since $a_1$ is a positive integer less than or equal to $a$.

Denote by $\sigma (a_n)$ the remainder of the division of $2019$ by the sum of the digits of $a_n$ in base $4038$.

We consider the sequence $\{b_n:\sigma (b_n)=0{\}}_{n=0}^{\infty }$ formed by the numbers such that their sum of digits in base $4038$ is divisible by $2019$. We have $b_0=0$, $b_1=2019$, etc.

For any positive integer $n$, the numbers $\{\sigma (2019n),\sigma (2019n+1),\dots ,\sigma (2019n+2018)\}$ form a complete remainder system modulo $2019$ (because $2019n,\dots ,2019n+2018$ differ only in the last digit). Therefore $$2019n\le b_n\le 2019n+2018.(1)$$ Let us consider arbitrary sequence $a_1,a_2,a_3,\dots$ which does not satisfy the condition, i.e. only finitely many of its terms $a_{i_1},a_{i_2},\dots ,a_{i_{\ell }}$ satisfy $\sigma (a_{i_j})=0$. Then there exists a positive integer $M$ such that $a_m\in \{b_n{\}}_0^{\infty }$ for every $m\ge M$. This means that for every positive integer $N$ we have $a_{M+k}\in \{b_n{\}}_0^{\infty }$, $k=0,\dots ,N$.

Assume that $a<2019$. Since the terms of the sequence are mutually distinct, we conclude that $$2019N\le b_N\le \underset{0\le k\le N}{\max }{a}_{M+k}\le a(M+N)\Rightarrow N\le \frac{aM}{2019-a},\forall N,(2)$$ which is a contradiction because of the choice of (fixed) $M$ and $a$. Therefore every $1\le a<2019$ gives a sequence with the required properties.

For $a\ge 2019$, the sequence $1,b_1,b_2,\dots$ satisfies the restrictions $a_n:=b_{n-1}<2019(n+1)\le a(n+1)$ for every $n$ and $\sigma (a_n)=0$ for every $n\ge 1$. Therefore, only the first term of this sequence has sum of digits in base $4038$ which is not divisible by $2019$. Therefore $a\ge 2019$ could not bring new solution.