**Short answer:** We can start at any number larger then $4$.

*Solution:* Instead of $2011$, suppose we are given any integer $\alpha$, and we want to know when we can reach $\alpha$. Suppose we start at $k$, and $k>4$. Then choose $a=\lfloor \frac{k}{2}\rfloor$ and $b=\lceil\frac{k}{2}\rceil$. (If $k$ is even, this is just $a=b=\frac{k}{2}$). Then because $k>4$, we will have that $ab>k$, so the sequence grows and by repeating this it will eventually be larger then $\alpha$. Now, we can decrease the sequence by $1$ at any point by choosing $a=1$, $b=k-1$, so once we have passed $\alpha$, we can just decrease by one at a time until we reach exactly $\alpha$.

For integers $k\leq 4$, notice that there is no choice of nonnegative $a$, $b$ such that $a+b=k$ and $a\cdot b>k$, so we are confined to this short interval.