Nice Problem. Assume otherwise. We call the neighbor to the right of $x$ as its friend. Also, if the friend of $x$ is more than $x$ then $x$ is in increasing position. Similarly, if the friend of $x$ is less than $x$ then $x$ is in decreasing position. Note that the odds, evens are also placed alternately. Since the friend of $1$ is obviously bigger than $1$, we get that $1$ is in increasing position. Hence, all odd numbers must be in increasing position while all even numbers are in decreasing position. What numbers can be a friend of $2009$? Only $2010$. So $2009,2010$ stick together in this order. What numbers can be a friend of $2007$? Only $2008,2010$. But $2010$ already has $2009$ to the left of it. Thus, $2007,2008$ stick together in this order. Do this again and again until we get $1,2$ stick together in this order. However, since $2$ is in decreasing position, its friend can only be $1$, which is a contradiction.