Well, that's my solution. If you know another solution please post it! First, suppose that there are $10^{9}$ consecutive numbers that are special. Among these numbers, there's one that is divisible by $10^9$. Then, it can be represented as $a.10^9$, where $a$ is a number with 11 digits, since any special number has 20 digits. Thus, the maximum is at most $10^9 - 1$. Now, observe that the $10^9 - 1$ numbers $10^{19}+ a, a=1,2,...,10^9-1$ are special. To prove this, suppose that for some $a=1,2,...,10^9-1$, $10^{19}+a$ is not special. Then, we can write $10^{19}+a=(10^9+x)(10^{10}+y)$, where $x, y \ge 0 $ are integers. But if $x >0$, we have $(10^9+x)(10^{10}+y) \ge (10^9 +1).10^{10}=10^{19}+10^{10} \Rightarrow a \ge 10^{10}$, an absurd, and if $y>0$, we have $(10^9+x)(10^{10}+y) \ge 10^9.(10^{10}+1)=10^{19}+10^{9} \Rightarrow a \ge 10^{9}$, another contradiction. Thus, $x=y=0$, which implies $a=0$, contradiction again. So, $10^{19}+ a, a=1,2,...,10^9-1$, are consecutive special numbers, showing us that the maximum number of consecutive special numbers is $10^9-1$.