No bigger than 1!

Let's look at the $1, 11, 111, 1111, \dots$ sequence. Let $x$ be the number we are talking about. Since the sequence is infinite long, and $x$ is finite, there will be two item in the sequence which make the same remainder when it is divided by $x$ (Let $a_n$ and $a_k$ be these numbers). So $|a_n-a_k|$ wil contain only $0$ 's and $1$ 's and it will be divisible by $x$ .