Let $2006a_i + 13$ be the number of times $i$ moves clockwise minus the number of times it moves counterclockwise. Roughly speaking, $a_i$ is the number of times $i$ goes around the circle. Each exchange moves one number clockwise and the other clockwise, so \[ \sum_{i=1}^{2006}2006a_i+13=0\\ \sum_{i=1}^{2006}a_i \equiv 1\mod{2} \] Since the $a_i$'s are integer, there must exist paired numbers $i$ and $j$ with $a_i \neq a_j$. This means that $j$ has moved around the circle at least once relative to $i$, and so must have passed $i$ as desired.

alternative solution let $c_i,a_i$ denote the times number $i$ moves clockwice and anticlockwice. then by Fubini Principle,$\sum_{i=1}^{2006}a_i=\sum_{i=1}_{2006}c_i$ if they can be divided into $1003$ pairs with two numbers in any pair unexchanged,then for these two numbers$i,j$,$a_i-c_i=a_j-c_j=13+2006k_{ij}$. hence $0=\sum_{i=1}^{2006}a_i-c_i=13*2006+4012\sum_{(i,j)}k_{ij}\equiv 2(mod 4)$,contradiction!

furthermore,I'll ask whether the total number of such moves is ODD?