Lets make a bipartite graph of columns and rows. We will put an edge between a column and a row if there is a board cell on intersection of the column and the row. Every awesome set of rooks will be a maximal matching in this bipartite graph and its famous that size of maximum matchings are continues. So between 2008 and 2010 there should be a maximal matching with size 2009. Sorry for my bad english.

Here is a proof that maximal matching sizes are continuous, as mentioned by X_emad_X . We will show that if there exists a maximal matching of size $x$, then either $x$ is the maximum size of any maximal matching, or there exists one of size $x+1$ too. Let $R$ be the set of rows and $C$ be the set of columns, and create a bipartite graph as described by the above. Let $(r, c)$ denote the edge between $r \in R$ and $c \in C.$ Then, suppose that $(r_1, c_1), (r_2, c_2), \cdots, (r_x, c_x)$ is the maximal matching of size $x$, where $r_1, r_2, \cdots, r_x \in R$ and $c_1, c_2, \cdots, c_x \in C.$ Suppose that there doesn't exist a maximal matching of size $x+1.$ Let $R' = R/ \{r_1, r_2, \cdots, r_x\}$ and $C' = C / \{c_1, c_2, \cdots, c_x\}.$ Call some $r_i$ (resp. $c_i$) pivotal if there is a $c \in C'$ (resp. $r \in R'$) so that $(r_i, c)$ (resp. $(r, c_i)$) is an edge. Lemma. At most one of $r_i, c_i$ is pivotal, for each $1 \le i \le x.$ Proof. If not, let $r \in R', c \in C'$ be so that $(r_i, c), (r, c_i)$ were both edges. Then, replacing $(r, c)$ with $(r_i, c)$ and $(r, c_i)$ gives us a maximal matching of size $x+1$, contrary to our assumption. $\blacksquare$ Let's now form the set $S$ as follows. For each $1 \le i \le x$, we will put whichever of $r_i, c_i$ is pivotal (there is at most one by the lemma) in $S$, and if none is pivotal just put one at random. Note that $|S| = x$, and that $S$ has the unique property that every edge in the graph has an endpoint in $S$. To see this, note that $(r_i, c_i)$ has an endpoint in $S$ for each $1 \le i \le x$, and for $c \in C'$, $(r_i, c)$ must have an endpoint in $S$ because $r_i \in S$ (since $r_i$ is pivotal). Similar argument holds for $(r, c_i).$ In this way, it's clear that no matching can have more than $x$ edges in it, because by Pigeonhole there would then exist two edges sharing a common endpoint in $S$. $\square$

Problem very hard