Let $n$ points be closer to $A$, and $(2009-n)$ points be closer $B$. So let the points that are closer to $A$ be $x_1,x_2, \dots , x_n$ and the ones closer to $B$ be $y_1,y_2, \dots , y_{2009-n}$. The sum of the distances from $A$ is $x_1 + x_2 + \cdots + x_n + y_1 + y_2 + \cdots + y_{2009-n} + (2009-n)(AB)$. From $B$, $x_1 + x_2 + \cdots + x_n + y_1 + y_2 + \cdots + y_{2009-n} + (n)(AB)$. So for them to be equal, FTSOC, we must have $(2009-n)(AB) = n(AB) \implies 2009 - n = n \implies 2009 = 2n$ which is clearly false.

Let $AB=n$ then suppose there are $m$ points closer to $A$ then $B$ (note that $m$ is a positive integer). Denote the sum of the distances from these points to $A$ as $X$. Then there are $2009-m$ points closer to $B$ than $A$. Denote the sum of the distances from these points to $B$ as $Y$. Then suppose the statement is true. We must have $$X+Y+m(n)=X+(2009-m)n+Y$$$$m=2009-m$$$$m=\frac{2009}{2}$$however this is a contradiction so we can't have the sums being the same.