Consider a sequence of numbers $a_1=1, a_{n+1}=10a_n+1$ $a_n=\underbrace{111...11}_{n}$ Set $S$ of residues $\pmod{1999}$ is finite, so it is easy to see that $a_n \pmod{1999}$ is periodic. $1\in S$, therefore, there are infinitely many numbers $a_n$ such that $a_n \equiv 1 \pmod{1999}$.

@above you still need to show that it can't be 1,2,2,2,2,2,2 etc or something similar (this is not difficult, though)

Emirhan wrote: Prove that, we find infinite numbers such that, this number writeable $1999k+1$ for $k \in {\mathbb N}$ and all digits are equal. It's application of Fermat Little's theorem. Apply F.L.T, we have $10^{1998} \equiv 1\pmod{1999} \implies 10^{1998L} \equiv 1\pmod{1999}$ So $1999\mid 99999\ldots 99 \implies 1999\mid 11111\ldots 11$ (because $\text{gcd}(9; 1999) = 1$). So $1999\mid 11111\ldots 110$. Then the number $11111\ldots 110 + 1 = 1999k + 1$ is the number that we finding . Check me if i made a mistake

BTW, i hope to see more elegant and nicer solution to this Can someone post a solution with CRT, please?