1. Let $x$ be the number of $+1$'s and $y$ be the number of $-1$'s among the 2002 numbers. Therefore, we have:
$$x + y = 2002$$

2. Consider the product of every two adjacent numbers. If two adjacent numbers are the same (both $+1$ or both $-1$), their product is $+1$. If they are different (one $+1$ and one $-1$), their product is $-1$.

3. Let $P$ be the sum of the products of all adjacent pairs. We are given that $P$ is negative:
$$P = \sum (\text{product of adjacent numbers}) < 0$$

4. To find $P$, note that there are $x$ pairs of adjacent $+1$'s, $y$ pairs of adjacent $-1$'s, and $2002 - (x + y) = 2002 - 2002 = 0$ pairs of adjacent $+1$ and $-1$. Therefore:
$$P = x \cdot (+1) + y \cdot (+1) + (2002 - x - y) \cdot (-1)$$
Simplifying, we get:
$$P = x + y - (2002 - x - y) = x + y - 2002$$

5. Since $x + y = 2002$, we substitute:
$$P = 2002 - 2002 = 0$$
This contradicts the given condition that $P < 0$. Therefore, our assumption must be incorrect.

6. Reconsider the problem. The correct approach is to count the number of $+1$ and $-1$ pairs. Let $a$ be the number of pairs $(+1, +1)$ and $b$ be the number of pairs $(-1, -1)$. Then:
$$P = a + b - (2002 - a - b)$$
Simplifying, we get:
$$P = 2(a + b) - 2002$$

7. Given $P < 0$, we have:
$$2(a + b) - 2002 < 0 \implies a + b < 1001$$

8. Since $a + b$ is the number of pairs of the same sign, the number of pairs of different signs is $2002 - (a + b)$. Therefore, the number of $+1$'s and $-1$'s must be balanced such that:
$$x \approx y$$

9. Let $S$ be the sum of the original numbers:
$$S = x - y$$
Since $x + y = 2002$ and $x \approx y$, we have:
$$|S| = |x - y| \leq 1000$$

The final answer is $\boxed{ |S| \leq 1000 }$