@MiloBrandt already posted a solution (make sure to upvote him!), but I wanted to give an alternative explanation (same idea).
As Milo noted, the number of zombies never decreases and therefore at some moment it stabilizes, say at $1 \leq K\leq 9$ zombies after time $T$.
Now consider any $1000$ consecutive zombies appearing past time $T$ and take a picture of the lawn in the moment each of them gets dropped on it - this makes a total of $1000$ pictures. Since on every picture there are exactly $K$ zombies, we see exactly $1000K$ zombies on all pictures.
Now notice that almost all of the selected $1000$ zombies appear on as many pictures as lawn spots they travel. The zombies for which this is not true are just the $K$ zombies, which appear on the last picture. We easily see that they have traveled between $1+2+...+K-1$ and $(10-K)+(11-K)+...+8$ spots more than the number of pictures they appear on.
Similarly, on the $1000$ pictures we have made, there are $K-1$ additional zombies, which appear multiple times (you can see all of them on the first picture). The total number of times they show up is again between $1+2+...+K-1$ and $(10-K)+(11-K)+...+8$.
Therefore we conclude that the total distance the selected $1000$ zombies travel is \begin{align}&&1000K \pm \{[(10-K)+(11-K)+...+8]-[1+2+...+K-1]\}\\ =&&1000K \pm (9-K)(K-1). \end{align} Since $(9-K)(K-1)$ is a number between $0$ and $16$, this solves the problem.
This problem was supposedly the hardest question from this year's IMO, but thought it could be translated into a nice puzzle. Hope you liked it. http://imo2015.org/solution.php?lang=en
Once again, make sure to upvote Milo's answer, as he uploaded first a legit solution.