Solution:

Suppose that $n$ of the original numbers are $+1$, so that the remaining $2002-n$ are $-1$, and their sum is
$$1 \cdot n + (-1) \cdot (2002-n) = 2n - 2002.$$
Also suppose that $m$ of the products equal $-1$; then the remaining $2002-m$ equal $1$, and the sum of the products is
$$1 \cdot (2002-m) + (-1) \cdot m = 2002 - 2m.$$
Since this is negative, $m > 1001$. Now, each product of $-1$ must come from a $+1$ and a $-1$ among the original numbers, and each $-1$ among the original numbers can contribute to at most two such products; hence the original numbers include more than $1001/2 > 500$ $-1$'s. Thus,
$$2002 - n \geq 501 \Rightarrow n \leq 1501 \Rightarrow 2n - 2002 \leq 1000$$
Similarly, the original numbers include more than $500$ $+1$'s, so
$$n \geq 501 \Rightarrow 2n - 2002 \geq -1000$$
Thus, $-1000 \leq 2n - 2002 \leq 1000$, which is what we need.