Knightwise Connectedness of a Chessboard Fragment Define the “Knightlyness” of a square on a chessboard fragment as the number of squares on that fragment that a knight could jump to in one move from the square under consideration. The “average Knightlyness” of a particular fragment is the arithmetic mean of all the Knightlynesses on that fragment. For the purposes of this problem, assume that all chessboard fragments are “strictly connected”, in the sense that, if two squares share a corner and do not both share an edge with a third square, then they share an edge. Determine whether, for all cases of integer $n$ from 0 to 7, inclusive, whether an average Knightlyness of $n$ is achievable.
I haven’t figured out any solution other than bashing. I’ve solved the cases for $n=0,1,2,3$, but I intuit that there should be something more elegant. I could be wrong, but I hope not- quite a bit of my honor lies on my being more correct than one of my classmates in a bet as to whether there is a nicer, more general solution (I made the bet because I wrote the question).
 A: I assume that chessboard fragments need not be parts of an $8\times 8$ chessboard, but can extend arbitrarily (are part of an artbitrarily large $n\times n$ chessboard)
In general we determine the average knightlyness by properties of a graph having the fields as vertices, and an edge between two vertices if they are a knight move apart. With $v$ vertices and $e$ edges, the average knightlyness (aka. average degree) is $\frac{2e}v$.
In a sufficiently large $n\times n$ chessboard, the average degree  will be arbitrarily close to $8$ as we have $O(n^2)$ fields of degree $8$ and only $O(n)$ fields of lower degree. 
Indeed, we readily see that for $n,m\ge 2$, an $n\times m$ chessboard results in 
$$ e=2(n-2)(m-1)+2(n-1)(m-2)=4mn-6(n+m)+8$$
edges (and of course $v=nm$ vertices).
In particular, the average degree of a $24\times 23$ board is
$$ 2\cdot \frac ev=2\cdot \frac{1934}{552}=7\frac1{138},$$
or just slightly above our highest target value $7$.
Now observe what happens when we add protrusions to our rectangle as in the following illustration:

A protrusion of height $1$ as on the left (at sufficient distance from the rectangle ends and from any other protrusion) adds $4$ to $e$ and $1$ to $v$.
Similarly, a protrusion of height $2$ adds $6$ to $e$ and $2$ to $v$.
Finally, a protrusion of greater height still adds $6$ to $e$, but an arbitrary amount $\ge 2$ to $v$.
Hence if we add two such protrusions of height $2$ to our $24\times 23$ board, we get $1946$ edges and $556$ vertices and hence a new average degree of
$$ 2\cdot \frac{1946}{556}=7.$$


Similarly, for a $12\times 11$ board, we have $e=398$, $v=132$, so $\frac{2e}v=6\frac1{33}$.
If we add one protrusion of height $1$ and one of height $2$, we arrive at $e=408$, $v=135$ and $\frac{23}v=6\frac 2{45}$ is still too big. But at least the numerator is a multiple of $6$, and all we have to do is raise the numerator to $136$, i.e., turn the height-2 protrusion into a height-3 one.

We repeat the game: For a $8\times 7$ board, $e=142$, $v=56$, $\frac{2e}v=5\frac1{14}$. Add one height-2 and two height-1 protrusions to obtain $e=160$, $v=60$. Turn the height-2 to a height-6 to finally arrive at $e=160$, $v=64$, $\frac{2e}v=5$.


Next: For $6\times 5$, $e=62$, $v=30$. Add a height-2 protrusion to obtain $e=68$, $v=32$. Increase $v$ to $34$, i.e., enlarge the protrusion to height $4$, and arrive at $\frac{2e}v=4$.


You seem alread have solved the cases of average degree $\le 3$, so I'll leave these out (and once can find nicer solutions for these by hand).
