$\color{blue} \textbf{\text{Answer.}}$ $\boxed{k=201}$. $\color{blue} \textbf{\text{Construction}}$ showing that $k=200$ does not work. We use a coordinate system such that pair $(a,b)$ denotes $a$-th column counted from the left to the right and $b$-th row counted from the bottom to the top. The rooks are placed, described by such function, where $x$ varies from $1$ to $300$ (which assures that the entire board is controlled by the rooks): \[y= \begin{cases} 300-\frac{x-1}{2} & \text{if } x\text{ is odd} \\ 300-\frac{x -2}{2} &\text{if } x \text{ is even.} \end{cases}\]Therefore, our rooks are placed on $(1,300),(2,300),(3,299),\ldots, (300,151)$. Consider $200\times 200$ square that has corner coordinates $(1,1)$, $(1,200)$, $(200,1)$ and $(200,200)$. Notice that $$y\leq 200\implies 200\geq 300-\frac{x-(1\text{ or }2)}{2}\implies x\geq 201$$and $$x\leq 200\implies y+\frac{200-(1\text{ or }2)}{2}\geq 300\implies y\geq 201,$$thus in fact no rook fall into such $200\times 200$ square. $\color{blue} \textbf{\text{Proof}}$ that $201\times 201$ always contains at least one rook. For the sake of contradiction assume that there exist a $201 \times 201$ square (call that square $\mathcal{S}$) that does not contain any rook. As rooks must control the entire board, rooks must also control this $201\times 201$ square. Assume that every column of $\mathcal{S}$ is covered by the rooks in the same column. Then as there are $99$ rows for such rooks, the maximal number of columns that can be covered is $99\cdot 2=198<201$, contradiction, therefore there exists a column in $\mathcal{S}$ that does not have rook in that column. Therefore, we must have every $201$ rows of $\mathcal{S}$ controlled by the rooks in those $201$ rows in $99$ columns. In the same manner, we have the maximal number of rows that can be covered, is $99\cdot 2=198<201$, therefore our initial assumption was wrong.

In general, if $300$ is replaced by $3n$ with $n \in \mathbb Z_{>0}$, then the answer is $\boxed{2n+1}$. So in this case of $n=100$, answer is $201$. First we give a construction for why $k=2n$ doesn't work. Below are the constructions for $n=3,4$, which easily generalize: [asy][asy] size(200); for(int i=1;i<4;++i){ fill(shift(i,10-i)*unitsquare,grey); fill(shift(10-i,i)*unitsquare,grey); } for(int i=4;i<7;++i){ fill(shift(i,i+3)*unitsquare,grey); fill(shift(i+3,i)*unitsquare,grey); } for(int i=1;i<11;++i){ draw((i,1)--(i,10)); draw((1,i)--(10,i)); } draw((1,1)--(7,1)--(7,7)--(1,7)--(1,1),red+linewidth(1)); [/asy][/asy] [asy][asy] size(250); for(int i=1; i<5;++i){ fill(shift(i,13-i)*unitsquare,grey); fill(shift(13-i,i)*unitsquare,grey); } for(int i=5; i<9;++i){ fill(shift(i,4+i)*unitsquare,grey); fill(shift(4+i,i)*unitsquare,grey); } for(int i=1;i<14;++i){ draw((1,i)--(13,i)^^(i,13)--(i,1)); } draw((1,1)--(9,1)--(9,9)--(1,9)--(1,1),red+linewidth(1)); [/asy][/asy] Now FTSOC assume some $(2n+1) \times (2n+1)$ square doesn't contain any rook. WLOG, this square is on the bottom left. [asy][asy] size(250); fill((1,13)--(10,13)--(10,10)--(1,10)--(1,13)--cycle,green+grey+grey); fill((13,1)--(13,10)--(10,10)--(10,1)--(13,1)--cycle,fuchsia+grey+grey); for(int i=1;i<14;++i){ draw((1,i)--(13,i)^^(i,1)--(i,13)); } draw((1,1)--(10,1)--(10,10)--(1,10)--(1,1),blue+linewidth(1)); [/asy][/asy] Let $A$ be the green square on top and $B$ be the pink square on right. By assumption, the blue square does not contain any rooks. Now $A$ is a $(2n+1) \times (n-1)$ rectangle. It must contain at most $2n-2$ rooks (otherwise, some three rooks in same row). So there cannot be a rook in each of the $2n+1$ rows. Particularly, there is some row $\mathcal R$ not containing any rooks. Similarly, looking at $B$ gives there is some column $\mathcal C$ not containing any rooks. But then the unit square $\mathcal R \cap \mathcal C$ is not attacked by any rook, contradiction. $\blacksquare$