Solution:

We may assume no entry of $A$ is zero, for otherwise we are done. We sort $A$ into a new list $B = (b_{1}, \ldots, b_{2000})$ by selecting elements from $A$ one at a time in such a way that $b_{1} > 0$, $b_{2} < 0$ and, for each $i = 2, 3, \ldots, 2000$, the sign of $b_{i}$ is opposite to that of the partial sum
$$s_{i-1} = b_{1} + b_{2} + \cdots + b_{i-1}.$$
(We can assume that each $s_{i-1} \neq 0$ for otherwise we are done.) At each step of the selection process a candidate for $b_{i}$ is guaranteed to exist, since the condition $a_{1} + a_{2} + \cdots + a_{2000} = 1$ implies that the sum of unselected entries in $A$ is either zero or has sign opposite to $s_{i-1}$.
From the way they were defined, each of $s_{1}, s_{2}, \ldots, s_{2000}$ is one of the 1999 nonzero integers in the interval $[-999, 1000]$. By the Pigeon Hole Principle, $s_{j} = s_{k}$ for some $j, k$ satisfying $1 \leq j < k \leq 2000$. Thus $b_{j+1} + b_{j+2} + \cdots + b_{k} = 0$ and we are done.