First notice that parity of thief's $X$ coordinate changes every time he moves.

Assume that at the beginning $X$ coordinate of thief's position is odd, and check districts $(1,1)$ and $(1,2)$. The next day check districts $(2,1)$ and $(2,2)$ and so on until ${1000}^{th}$ when you check districts $(1000,1)$ and $(1000,2)$. What is achieved this way is that if starting parity was as assumed, thief could have never moved to district with $X$ coordinate $i$ on day $i+1$, hence he couldn't have jumped over the search party and would've been caught. If he wasn't caught, his starting parity was different than we assumed, so on ${1001}^{st}$ day we search whatever ($1$ and $1001$ are of the same parity, so we need to wait one day), and then starting on ${1002}^{nd}$ day we do the same sweep from $(1,1)$ and $(1,2)$ to $(1000,1)$ and $(1000,2)$ and guarantee to catch him.

Shortest possible solution is by going from $(2,1)$ and $(2,2)$ to $(1000,1)$ and $(1000,2)$ twice in a row, a total of $1998$ days, which is correct in the same way. First sweep catches the thief if he started with even $X$ coordinate, and second sweep catches the thief if he started with odd $X$ coordinate.