Very nice problem! Note that a square's value is the number of ways one can get from the top left square to that square only using right and down movements but not traversing through any marked squares. Remark that if no squares were marked, then the bottom right corner would be $\dbinom{4022}{2011} \equiv 2 \pmod{2011}$. Now, it is easy to see that marking the square $(a, 2011 - a)$ (the first coordinate denotes $a$ moves to the right, second that many moves down) decreases the number of ways by $\dbinom{2011}{a}^2$ which is divisible by $2011$ since marked squares cannot be the corners. It immediately follows no matter which squares you mark off the bottom right is still $2 \pmod{2011}$, so the result follows. Remark: This generalizes to $(p+1) \times (p+1)$ board well when $p$ is prime, in that the bottom right square is $2 \pmod{p^2}$.

correct me if i'm wrong please but isn't the bottom-right number mod $2011$ the sum of the top-right and the bottom-left number with the only need for the rule to apply for numbers below the bottom-left top-right diagonal(btw how is that diagonal called).