Another approach: Let $T_n = \sum_{A \in S_n} PA^2$ where $P$ is a fixed point and $S_n$ is the set of points after $n$ operations. Note that $T_n \geq 0$. We claim that the sequence $T_n$ is strictly decreasing. This is true since $$PA^2 + PB^2 = 2(MA^2 + MP^2) \geq 2(PM^2 + CM^2) = PC^2 + PD^2$$by Apollonius theorem and the condition that $AB \geq CD$. Thus, there can be only finitely such operations performed .

Set an origin arbitrarily. Over all points $P$ in the set, we claim $\sum|P|^2$ is strictly decreasing, which finishes. Indeed, say that when we replace $A,B$ with $C,D$, we have a common midpoint $M$. Then, $A=M+x$, $B=M-x$, $C=M+y$, $D=M-y$ for some $|x|>|y|$. Then, \[|A|^2+|B|^2-|C|^2-|D|^2=(M+x)^2+(M-x)^2-(M+y)^2-(M-y)^2,\]which is equal to $2|x|^2-2|y|^2>0$, as desired.

We claim that the value $\sum_{AB\in S}(AB)^2$ strictly decreases. This obviously finishes. Note that we obviously have $(AB)^2>(CD)^2$. Next, note that by Appolonius theorem , we have that for any point $P\in S$ (Denote by $M$ the midpoint of $AB$.) : $$PC^2+PD^2-\frac {CD^2}{4} = PM^2 = PA^2+PB^2-\frac {AB^2}{4} \implies PC^2+PD^2 = PA^2+PB^2 - \underbrace {\left( \frac {AB^2}{4}- \frac {CD^2}{4}\right)}_{>0}$$ Hence the claim holds and we are done. $\blacksquare$

Consider some point $P$ far away from $S$. We claim that the quantity $\sum{Q\in S} |PQ|^2$ increases. This value must always be an integer, so this clearly finishes. To prove, this we use Stewart's Theorem. Specifically, if $|AB|=2m$ and $M$ is the midpoint of $AB$, then \[PA^2 + PB^2 = 2m^2 + 2\cdot |PM|^2 = \frac12 |AB|^2 + 2\cdot |PM|^2\]Note that the midpoint of $CD$ is also $M$, so \[PC^2+PD^2 = \frac12 |CD|^2 + 2\cdot |PM|^2 < frac12 |AB|^2 + 2\cdot |PM|^2 = PA^2+PB^2\]so this value always decreases and we are done. $\blacksquare$.

Oh I did this problem ages ago and never wrote it up I claim that the sum of $AB^2$ over all $A,B \in \mathcal{S}$ strictly decreases. Indeed, in any operation the midpoints of $\overline{AB}$ and $\overline{CD}$ are the same—call this point $M$. Then for any point $P \in \mathcal{S}$ other than $A,B,C,D$, by the median length formula we have $$AB^2>CD^2 \implies 2PA^2+2PB^2-PM^2>2PC^2+2PD^2-PM^2,$$which, upon summing over all $P$ and using $AB^2>CD^2$ one more time implies the desired claim. To finish, note that since $\mathcal{S}$ is a set of lattice points, the sum of $AB^2$ cannot strictly decrease an unbounded number of times. $\blacksquare$