National Olympiad of Argentina

The operation replaces $a$ and $b$ by $|b-a|$ which is even if $a$ and $b$ have the same parity and odd otherwise. So the number $N$ of odd numbers either remains unchanged or decreases by $2$ after each step. Initially $N$ is odd ($N=1005$), so the last number will be odd, and clearly between $1$ and $2010$. Conversely each odd number in this range can end up as the last one after a sequence of operations. Let $2k-1$ be such a number, $1\le k\le 1005$; then $2k\le 2010$. Separate the pair $(1,2k)$ and divide the remaining numbers into $1004$ pairs of consecutive integers: $$(2,3),(4,5),\dots ,(2k-2,2k-1);(2k+1,2k+2),\dots ,(2009,2010).$$ Apply the operation to each of these pairs to obtain $1004$ ones. They can be grouped in $502$ pairs, and each of these yields a zero. The last pair $(1,2k)$ gives $2k-1$. So we obtain $2k-1$ and several zeros, after which it is clear that the last number will be $2k-1$.