Consider the sum of the squares of the $2000$ integers on each turn.

Consider the sum of the squares of any $2000$ consecutive integers. Let $a$ be an arbitrary integer. Then $\sum_{i=1}^{2000}(a + i) \equiv \sum_{i=1}^{2000}i \equiv \frac{2000 \times 2001 \times 4001}{6} \equiv 1000 \pmod {2000}$ Now whenever we use a turn, we turn $(a, b)$ into $(a + b, a - b)$. The sum of the squares of the initial pair is equal to $a^2 + b^2$. But the sum of the squares of the new pair is equal to $(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)$ which means that the sum of the squares of all of the $2000$ integers we have will always double each turn. Considering the sum of the squares mod $2000$, we are trying to obtain $1000 \to 1000$. But $1000 \to 0 \to 0 \to ...$ since the sum of the squares always double each turn. Therefore it is impossible to write $2000$ consecutive integers again. Q.E.D