The answer is $n = 11$. The $11$ integers $2^0, 2^1, \dots, 2^{10}$ certainly work. Now we show that $n = 10$ does not work.

Suppose on the contrary that the $10$ integers $x_1 < x_2 < \dots < x_{10}$ satisfy the requirement. They can form $2^{10} = 1024$ sums, not necessarily distinct. There are $1001$ integers from $1000$ to $2000$. So at most $23$ of these sums can be $< 1000$. Now $x_1, \dots, x_9$ can pair up to form $\binom{9}{2} = 36$ sums. So at least one of the sums is $\ge 1000$. Therefore $x_9 > 500$.

Let $s$ be the sum of some of the integers in $x_1, \dots, x_8$. Since $x_9 + x_{10} \ge 1003$, at most one of $s \ge 1000$ and $s + x_9 + x_{10} \le 2000$ can hold. Therefore at most $3$ of the $4$ sums
$$s,\ s+x_9,\ s+x_{10},\ s+x_9+x_{10}$$
can be between $1000$ and $2000$ (both inclusive). The integers $x_1, \dots, x_8$ can form at most $2^8$ different sums. Therefore $x_1 \dots x_{10}$ can form at most $3 \times 2^8 < 1000$ sums between $1000$ and $2000$. So $n = 10$ is impossible.