We first notice the following crucial fact:
The number of zombies on the field never decreases.
This is obvious since each step adds one zombie and removes at most one. However, since the number of zombies can never exceed $9$, it follows that, after some time $T$, the number of zombies on the field is constant*. We will prove that zombies created after time $T$ satisfy the desired condition. After this point, the following rule must be observed:
If there is not a zombie at the leftmost plot, a zombie must be added there.
We must follow this as otherwise one would be adding a zombie without removing one.
Now, we consider that, if there are $n$ zombies on the field at all time, we would expect that, as they walk a total of $n$ steps left each turn. We get a bit of a hint on how to use this fact in the $n=2$ case - in particular, if we make a list of which plot the zombies are placed at (with $1$ being the leftmost), we might see a pattern like:
$$2,2,3,1,5,1,1,1,4,1,1,2,2,\ldots$$
where (after a point) we will necessarily see a concatenation of patterns of the form:
$$k,\underbrace{1,1,\ldots,1,1}_{k-2\text{ times}}$$
since we have to "wait" for the zombie placed at $k$ to reach the start. When we're summing over such a pattern, we can see that we can "shift" the mass away from the $k$ and towards the $1$s to leave a sum of $2(k-1)$. In a sense, when we put a $k>2$ in the pattern, we're taking a "loan" from later in the pattern, which we slowly repay by taking only $1$ step (less than the "allotted" $2$ steps per zombie).
This suggests to us we want to find some function which takes a state of the game as input and tells us how much "debt" we have to repay (we might also call it an energy function). We will define a state as a subset $S$ of the squares in the garden, representing the set which are occupied by zombies right before a new one is placed. Note that thus the rightmost square will never be occupied, as a zombie placed there immediately wanders left and that if there are $n$ zombies $|S|=n-1$ since the state occurs after a zombie has just walked off and before its replacement is added. A legal transition from $S$ to another state consists adding a new element, then decrementing all the elements and removing $0$ if it appears. We wish to find a function $E_n(S)$ on states such that $|S|=n-1$ which satisfies the following crucial property:
Suppose state $S'$ could occur immediately after state $S$ when Dr. Zomboss adds a zombie at position $k$. Then $E_n(S)-E_n(S')=n-k$.
This is to say we are measuring the difference between a given move and the allotment of zombie steps - so if we've got "too many" steps $E_n(S')$ will be more than $E_n(S)$. Notice that this defines $E_n$ up to a constant factor if it exists at all. One may explicitly compute it by choosing $E_n(S)=e$ for some arbitrarily chosen $S$ and $e$ and finding the energy of all states reachable from it - and thinking about that idea for a second leads us to the following valid energy function:
$$E_n(S)=\sum_{s\in S}s$$
One may check that this satisfies the desired condition.
Then, the rest is easy: Suppose $S_0$ was the state before the first zombie included in the sum is added and $S_{1000}$ was the state after the last zombie in the sum was added (and the intervening $S_i$ are defined accordingly). In particular, we may show that the number of steps walked by those $1000$ zombies is:
$$E(S_{1000})-E(S_0)+1000n.$$
This is easy because if the $i^{th}$ zombie walked a distance of $k_i$, then $E(S_{i})-E(S_{i-1})=k_i-n$, thus we may algebraically manipulate the desired sum as in the following sequence of equivalent expressions:
$$k_1+k_2+k_3+\ldots+ k_{999}+k_{1000}$$
$$(k_1-n)+(k_2-n)+(k_3-n)+\ldots + (k_{999}-n)+(k_{1000}-n)+1000n$$
$$[E(S_1)-E(S_0)]+[E(S_2)-E(S_1)]+\ldots +[E(S_{1000})-E(S_{999})]+1000n$$
where this last sum telescopes to
$$E(S_{1000})-E(S_{0})+1000n.$$
Now, if we let $A$ be the state with the least energy and $B$ be the state with the most, the we have:
$$E(A)-E(B)+1000n \leq E(S_{1000})-E(S_{0})+1000n\leq E(B)-E(A)+1000n.$$
meaning the difference between any pair of such sums may be at most $2[E(B)-E(A)]$.
Clearly, $B=\{8,7,\ldots,9-n\}$ has the most energy and $A=\{1,2,\ldots, n\}$ has the least - and $A$ is a translation of $B$ leftwards by $(8-n)$ meaning the difference between the energy is $n(8-n)$. This is maximized when $n=4$ yielding $16$ as the total difference in energy. Thus, no such sum may differ by more than $2\cdot 16=32$ after the field settles into a constant number of zombies.
(*More strongly, if we consider a game state to be the set of plots on which zombies lie and say $A\sim B$ if state $A$ could occur after state $B$ and vice versa, then it happens that $A\sim B$ is equivalent to $|A|=|B|$ meaning that the number of zombies is essentially the only invariant of this type - so we're in trouble if this $T$ doesn't cut it!)
