Let $a_t$ be a number whose decimal expansion has $t$ $1$ digits, e.g. $a_7=1111111$. Note that for $i>j$, $a_i-a_j=a_{i-j}*10^j$, e.g $a_5-a_2=11111-11=11100=a_3\times 10^2$.
Proposition:
Let $n=2m$ where $m$ is any odd number not divisible by $5$. Then among the numbers $nk+1$, k=1,2,3,... there are infinitely many $a_t$ numbers.
For $m=999, n=1998$ this proposition answers the question.
Proof:
The set of remainders of the numbers $a_t$ by $m$,
$$R=\{a_t \mod m : t=1,2,3,...\}$$
is a finite set (whose cardinality is at most $m$), and so there exists $T$ such that
$$R=\{a_t \mod m : t=1,2,3,..,T \}$$
Now take the numbers $a_{2T},a_{3T},a_{4T},..$. For each $a_{iT}$ there exists $a_{s_i}$, $1<=s_i<=T$, such that $a_{iT} \equiv a_{s_i} \mod m$. Therefore $m$ divides $a_{iT}-a_{s_i}$. But
$$a_{iT}-a_{s_i}=a_{iT-s_i} \times 10^{s_i}$$
Since neither $2$ nor $5$ divide m, it follows that m divides $a_{iT-s_i}$, and thus $n=2m$ divides $a_{iT-s_i} \times 10$. So for some $k_i$
$$a_{iT-s_i} \times 10 + 1 = n\times k_i +1 $$
But $a_{iT-s_i} \times 10 + 1$ is just $a_{iT-s_i+1}$. So for each $i = 2,3,..$, there exist $s_i \in {1,..,T}$, and $k_i$ such that
$$a_{iT-s_i+1} = nk_i + 1$$
Since for any $i \neq j$ we have $a_{iT-s_i+1} \neq a_{jT-s_j+1}$, this proves the proposition.