This is a question that is tripping me up- for $A$, I've it such that each term in the square bracket is expressed as $\frac{(2010-2a+1)2010!}{a!(2010-a+1)!}$, where $a = 0, ..., 1005$.
However, I am unsure how the square will affect the term, and in particular, what is the relationship between the sum of square of this term and that of $\binom{4020}{2010}$.
Any help will be greatly appreciated, thank you!
We will make use of the following identity: $$ \sum_{k = 0}^{a} \binom{a}{k} \binom{b}{m - k} = \binom{a + b}{m}. $$
This can be proven by counting the number of ways to select $m$ objects from a set of $a + b$ objects. This is clearly equal to $\binom{a + b}{m}$. On the other hand, if we choose $k$ objects from the first $a$ objects, then there are $\binom{a}{k}$ ways to choose those objects, and $\binom{b}{m-k}$ ways to choose the remaining objects.
In particular, we note that $$ \sum_{k = 0}^{2010} \binom{2010}{k}^2 = \sum_{k = 0}^{2010} \binom{2010}{k} \binom{2010}{2010 - k} = \binom{4020}{2010} $$ and $$ \sum_{k = 0}^{2010} \binom{2010}{k} \binom{2010}{k-1} = \sum_{k = 0}^{2010} \binom{2010}{k} \binom{2010}{2011 - k} = \binom{4020}{2011}. $$
We thus have that $$ A = \sum_{k = 0}^{1005} \left( \binom{2010}{k} - \binom{2010}{k - 1} \right)^2 = \sum_{k = 0}^{1005} \left( \binom{2010}{k}^2 + \binom{2010}{k-1}^2 - 2\binom{2010}{k} \binom{2010}{k-1} \right). $$
Now note that $$ \sum_{k = 0}^{1005} \binom{2010}{k - 1}^2 = \sum_{k = 1}^{1005} \binom{2010}{2011 - k}^2 = \sum_{k = 1006}^{2010} \binom{2010}{k}^2 $$ so that $$ \sum_{k = 0}^{1005} \left( \binom{2010}{k}^2 + \binom{2010}{k-1}^2 \right) = \sum_{k = 0}^{2010} \binom{2010}{k}^2 = \binom{4020}{2010}. $$
Similarly, $$ \sum_{k = 0}^{1005} \binom{2010}{k} \binom{2010}{k - 1} = \sum_{k = 1}^{1005} \binom{2010}{2010 - k} \binom{2010}{2011 - k} = \sum_{k = 1006}^{2010} \binom{2010}{k} \binom{2010}{k - 1} $$ so that $$ 2\sum_{k = 0}^{1005} \binom{2010}{k} \binom{2010}{k - 1} = \sum_{k = 0}^{1005} \binom{2010}{k} \binom{2010}{k - 1} + \sum_{k = 1006}^{2010} \binom{2010}{k} \binom{2010}{k - 1} = \sum_{k = 0}^{2010} \binom{2010}{k} \binom{2010}{k - 1} = \binom{4020}{2011}. $$
It follows that $$ A = \binom{4020}{2010} - \binom{4020}{2011} = \frac{4020}{2010} \binom{4019}{2009} - \frac{4020}{2011} \binom{4019}{2010} = \frac{2}{2011} \binom{4019}{2009}. $$
Thus $$ sA \geq \binom{4020}{2010} \iff \frac{2s}{2011} \binom{4019}{2009} \geq \frac{4020}{2010} \binom{4019}{2009} \iff s \geq 2011. $$
We see that the minimum possible value for $s$ is $2011$.