For $n=1$, we have $\sum_k \sum_l$ ${2k+2l}\choose{2l}$ ${4n-2k-2l}\choose{2n-2l}$ $= \sum_l$ ${2l}\choose{2l}$ ${4-2l}\choose{2-2l}$ + ${2+2l}\choose{2l}$ ${2-2l}\choose{2-2l}$ = ${4}\choose{2}$ + ${2}\choose{2}$ + ${2}\choose{0}$ + ${4}\choose{2}$ = $14$.

Forgot about $2$ terms being on both ends of a progression, so really they may form $9$ progressions, and that's a bad example. A better choice is N=13, where there is a set of size 7 with no progressions, but the smallest maximal set is $'\\{3,5,8,10\\}'$, and this time a set of $3$ elements can't be maximal, by the above argument.

An example of a set without 3-AP will be an upper bound for "smallest maximal subset", where maximal means no other numbers can be added without forming a 3-AP. For example, if $N=10$, then $\\{1,3,4,9,10\\}$ is a set with no 3-AP, and is also maximal, but it is not the smallest such set. $\\{2,3,5,6\\}$ has $4$ members, which is minimal, since $3$ elements can only form $6$ progressions (two for each pair), so at least one more element can be added.

I need to revisit intro to calculus and fix my awful notation: differentiated $f^2(x)$ instead of $f(f(x))$.

Hardy-Littlewood-Polya's "Inequalities" is an excellent source for proving these types of statements. You will want to check out sections 3.4 and 3.15 on mean values of arbitrary functions. Be forewarned, the book has more theorems than pages.

I suggest editing the tag (inf)infinite sequences, unless you are making a nod towards $\mathbb{N}$ being the smallest of infinities.

You may want to begin at $0$ in $r_{total}$ if it is meant to be all possible subsets of $n^2$ elements, otherwise it will be $2^{n^2} - 1$.

That won't work either. If $a=2$ and $b=3$, there is no way to partition $(1,2,3,4,5,6)$ into two sets of equal size and equal sum (both sums would have to be $10.5$), but there are plenty of ways to travel across a $6$ by $6$ chessboard and cut it into two pieces of $18$ squares each. Perhaps it is $a=2$, and an $\frac{n}{2}$ by $(n+1)$ chessboard.