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The sentence means that $(a,b)$ is a root of the $f$ with multiplicity at least $r$, with $r>(k+\ell)$. Therefore $f(a,b)=0$, $f'(a,b)=0$, $\cdots$, $f^{(r-1)}(a,b)=0$. Not sure about that conclusion, I haven't read the paper.
I think you might want to add some restrictions (such as connected subgraph) otherwise by pigeon-hole principle isn't this just $$ N = \max \left[ \min\left\{ k,\ m\leq \left\lceil \frac{1}{r}\binom{k}{2}\right\rceil \right\} ,n \right]$$ ?
Note that I would personally write the problem in term of independence number : If $G_1,\ldots, G_n$ are the graphs induced by a maximal coloring (sometime referred as a color Pattern of $K_M$), your question ask for the minimum independence number of $G_1$, over all maximum coloring. I find it this formulation easier to generalize.
My intuition would be "no" : I agree that the graph induced on $K_N$ by the colors $2,\ldots,n$ must be very close to be Ramsey for $(K_{a_2},\ldots, K_{a_n})$, but there are such graphs with small clique number (actually with max clique of size $K_{a_n}$ given a result by Folkman.). So I would expect that you might find "maximal coloring" without such set of vertices
My intuition is that it might be very difficult (if possible at all) to impose a constraint on the deleted edge such that you always encounter a Hamilton cycle. Make me think about some kind of Maker-Breaker game, you play against an opponent, but you can restrict its choices. There might be some intuition to gather from these, but it's not directly related.
What do you mean "encounter" a Hamilton cycle ? That the cycle cover $C$ must be Hamiltonian at one point ? And can you confirm that you are talking about "vertex cycle covers", it seems that edge and vertex covers are both sometimes called "covers". Then surely not, select one vertex $v$. If at each second step, you always delete an edge adjacent to $v$, then you can always find a cycle cover where e.g. $v$ is in a triangle $T$, and all vertices of $G-T$ are in one large cycle (because $G-T$ is complete), up to the point where $v$ has degree $1$, so there is no more cycle cover of $G$.
