The write-up for this problem is quite annoying.

Actually problem is easy, and it is not hard to find answer(similar pr can be found on IMO SL 2000). Mine answer is same as previous one's, and yes, $2k-100 \times 2k-100$ is not reachable. And If k is odd, reachable coordinates $(a,b)$ must have a and b have same parity, and that other cells with same parity coordinates are reachable can be proven easily by induction by denoting $n=100$, same thing works for $k$ even (this time other all cells are reachable).

The answer is $$\boxed{L(k)=\begin{cases} 4k(100-k),& k\equiv 0\pmod{2}\\ 2k(100-k),& k\equiv 1\pmod{2} \end{cases}}$$ Claim: $(i, j)$ with $i, j\in[101-k, k]$ is not reachable. Proof. For the sake of contradiction, say it was reachable. In that case, there would be a path from $(i, j)$ to $(1,1)$ too. However, there are no moves for the $k$-knight from that position without going out of bounds, a contradiction. $\blacksquare$ Delete the square formed by the set of cells $(i, j)$ with $i, j\in[101-k, k]$: This region is subsequently considered out of bounds. Note that the number of deleted cells is $(2k-100)^2$. Lemma: $(i, j)$ is reachable iff $(i-2, j)$, $(i+2, j)$, $(i, j-2)$ or $(i, j+2)$ is reachable, given that $(i, j)$ is within the bounds. Proof. If $(i, j)$ is reachable, at least one of $(i-2, j)$, $(i+2, j)$, $(i, j-2)$ and $(i, j+2)$ must be within the bounds as well; so suppose the cell $(i-2, j)$ is within the bounds. Either $(i-1, j-k)$ or $(i-1, j+k)$ is always within the bounds, so the $k$-knight goes along the path $(i, j)\to(i-1, j\pm k)\to(i-2, j)$. This shows it is reachable, and we can use similar arguments for the other cells. The converse is implicit. $\blacksquare$ Let the parity of $i+j$ be the parity of the cell $(i, j)$. Case 1: $k\equiv 0\pmod{2}$. It is possible to reach any $(i, j)$ within the bounds from $(1,1)$ such that $i\equiv j\equiv 1\pmod{2}$. After covering all such cells, move the $k$-knight from $(1,1)$ to $(k+1, 2)$. Since $k+1\equiv 1\pmod{2}$ and $2\equiv 0\pmod{2}$, it is possible to reach any $(i, j)$ within the bounds such that $i\equiv 1\pmod{2}$ and $j\equiv 0\pmod{2}$. After covering all such cells, move the $k$-knight from $(1,1)$ to $(2, k+1)$. Since $2\equiv 0\pmod{2}$ and $k+1\equiv 1\pmod{2}$, it is possible to reach any $(i, j)$ within the bounds such that $i\equiv 0\pmod{2}$ and $j\equiv 1\pmod{2}$. Since all odd cells within the bounds are reachable, the $k$-knight can go from $(1,1)$ to $(k+2, 1)$ and then to $(2,2)$. Since $2\equiv 0\pmod{2}$, it is possible to reach any $(i, j)$ within the bounds such that $i\equiv j\equiv 0\pmod{2}$. Therefore, we covered every cell within the bounds, and the number of such cells is $$100^2-(2k-100)^2=4k(100-k),$$as claimed. $\blacksquare$ Case 2: $k\equiv 1\pmod{2}$. Claim: Odd cells are not reachable. Proof. The parity of reachable cells is invariant. If at any point the $k$-knight is at $(i, j)$, then on the next move it would go to a cell of the same parity, because $i+j\pm k\pm 1\equiv i+j\pmod{2}$. Since we start at $(1,1)$, an even cell, it follows that a reachable cell must be even. $\blacksquare$ It is possible to reach any $(i, j)$ within the bounds from $(1,1)$ such that $i\equiv j\equiv 1\pmod{2}$. After covering all such cells, move the $k$-knight from $(1,1)$ to $(k+1, 2)$. Since $k+1\equiv 2\equiv 0\pmod{2}$, it is possible to reach any $(i, j)$ within the bounds such that $i\equiv j\equiv 0\pmod{2}$. Therefore, we covered all the even cells within the bounds, and the number of such cells is $$\frac{100^2-(2k-100)^2}{2}=2k(100-k),$$as claimed, and we are done. $\blacksquare$

Let's make some observations: 1.If $(a,b)\to (c,d)$ then $(c,d)\to (a,b)$ 2.If $101-k\le a,b \le k $ then $(a,b)$ is not reachable because WLOG we add/subtract $k$ from $b$ we see that $b-k\le 0$ and $b+k \ge 101$ so is not in the square. 3. If $k$ is odd we can see by induction that $a\equiv b \pmod 2$ This is enough for the bound now it remains to give the construction and final answer: Case 1: k is even We claim that $(a,b)$ is reachable as long as it doesn't satisfy 2.We will do this by strong induction on $a+b$ with a trivial base case: If $max(a,b)\ge k+1$ it is easy: WLOG $a\ge k+1$ and we have that $(1,1)\to (a-k,b+1)\to (a,b)$ and we are done. Otherwise $a,b\le k$ and WLOG $b\le 100-k$.If $a\ge 3$ we have $(1,1)\to (a-2,b)\to (a-1,b+k)\to (a,b)$ and we are done. Otherwise $a\in \{1,2\}$ and if $b\ge 3$ we have $a\le 100-k$ and do similar as above. It remains to check $(1,2)$ and $(2,2)$ $(1,1)\to (2,k+1)\to (k+2,k+2)\to (k+1,2)$ and now subtact $1$ form the $x$ coordinate and alternate $+k$ and $-k$ on the other coordinate.Since $k$ is even you will get $(1,2)$. $(1,1)\to (2,k+1)\to^{(*)} (2+k,2)\to^{(**)} (2,2)$ for (*) add and subtract $k, k-1$ times and since odd it will be $2+k$ and on the other coordinate subtract $1$ and you will get $2$ and for (**) subtract $1 k$ times and on the other alternate $+k$ and $-k$ Case 2: $k$ is odd Do the exact same induction step as in case 1 and you only have to check $(2,2)$ which is easy Now the final answwer is $\boxed{100^2-(2k-100)^2}$ for k even and $\boxed{2\cdot 50^2-2\cdot (k-50)^2}$ for k odd

This problem must be a joke. I think it's unsuitable for a math olympiad, more like an exercise to train students to write a proof rigorously. The answer is not hard to guess, but the write-up is a total pain. I thought the write-up of ISL 2007 C1 was already bad, but this one is a disaster!