Can you elaborate on "Remark 2"? Is this just for $a=1$ or is it really for any constant $a$?

Perhaps a bit more generally, if the average degree (i.e. average number of sets each element is contained in) of the hypergraph is constant, then $\min |A_1\cup \dots \cup A_n|=\Omega(\sum_{i=1}^n|A_i|)$.

A small observation is that if $a=0$, then every element is in at most two sets and thus $|A_1\cup \dots \cup A_n|\geq \frac{1}{2}\sum_{i=1}^n|A_i|$. So in this case $\min |A_1\cup \dots \cup A_n|=\Omega(n^2)$ even if the sets have much fewer than $n/3$ elements.

Spirograph en.wikipedia.org/wiki/Spirograph

Another observation is that you can assume $R=\emptyset$ since if there were a counterexample with $R\neq \emptyset$, then you could add all edges inside the set $R$ and all edges between the sets $A$ and $R$ which has at least as many pairs of vertices with degree difference at least $k$.

A couple of trivial comments: (i) You can let $B:=\{v_i:\deg(v_i)\leq n-k-1\}$ since there can be no vertices of degree $n$. (ii) Since $\mathcal{I}_k(G)\leq |A||B|$, you can assume you are in the case where $n-k<\min(|A|, |B|)\leq \max(|A|, |B|)<k$.

Note that you have asked this question before both explicitly mathoverflow.net/questions/370950/… and essentially mathoverflow.net/questions/266511/…

I just read the fantastically titled "Hedgehogs are not colour blind" paper. This is a very nice example with such a short, simple proof. Based on what you say in the abstract, there won't be a result in the literature which already answers my question (and thus it seems unlikely that my question has an easy answer).

Well, the minimum vertex cover is at least half the length of a longest path. So if the vertex cover number is small, then so is the length of the longest path. But as pointed out with the subdivided star, having large vertex cover number doesn't mean you have a long path

I see. Can a suggest a alternate phrasing: You are looking to minimize the largest element in a set $S$ of $\binom{n}{2}$ non-negative integers having the property that for every bijection $w:E(K_n)\to S$ and every pair of distinct vertices $u,v\in V(K_n)$, we have $\sum_{e\ni v}w(e)\neq \sum_{e\ni u}w(e)$.

I see. When you said "a set of $\binom{n}{2}$ integers" I wrongly interpreted that as "at most $\binom{n}{2}$ integers." So just to clarify, you want an irregular edge labeling with distinct non-negative integers and you want to minimize the largest weight?

Sure, I can do that. Perhaps after you dig into it, you can comment on my answer whether the 1-2-3 conjecture is known for complete graphs. I did see a recent paper which shows that for regular graphs the set {1,2,3,4} suffices. "The 1–2–3 Conjecture almost holds for regular graphs" by Pryzbylo