Solution:

For $n \leq 10$ Ann wins by writing the numbers $1, 2, \ldots, 2^{n-1}$. Indeed, the result Ivo can get is a non-zero integer between $-1023$ and $1023$, since it has the same sign as the largest remaining number $\left(2^{j} > 2^{j}-1 = \sum_{i=0}^{j-1} 2^{i}\right)$.

For $n \geq 11$ the set $C$ of Ann's numbers has $2^{n}-1 > 2003$ different nonempty subsets. Hence the sums of numbers of two of them, say $A$ and $B$, are congruent modulo $2003$. If Ivo puts $+$ in front of the numbers of $A \setminus B$, $-$ in front of the numbers of $B \setminus A$ and deletes the remaining numbers of $C$, he wins.