8. Let $S$ be the disk with the smallest radius, say $s$, and $O$ the center of that disk. Divide the plane into 7 regions: one bounded by disk $s$ and 6 regions $T_{1}, \ldots, T_{6}$ shown in the figure. Any of the disks different from $S$, say $D_{k}$, has its center in one of the seven regions. If its center is inside $S$ then $D_{k}$ contains point $O$. Hence the number of disks different from $S$ having their centers in $S$ is at most 2002. Consider a disk $D_{k}$ that intersects $S$ and whose center is in the region $T_{i}$. Let $P_{i}$ be the point such that $O P_{i}$ bisects the region $T_{i}$ and ! $O P_{i} = s \sqrt{3}$. We claim that $D_{k}$ contains $P_{i}$. Divide the region $T_{i}$ by a line $l_{i}$ through $P_{i}$ perpendicular to $O P_{i}$ into two regions $U_{i}$ and $V_{i}$, where $O$ and $U_{i}$ are on the same side of $l_{i}$. Let $K$ be the center of $D_{k}$. Consider two cases: (i) $K \in U_{i}$. Since the disk with the center $P_{i}$ and radius $s$ contains $U_{i}$, we see that $K P_{i} \leq s$. Hence $D_{k}$ contains $P_{i}$. (ii) $K \in V_{i}$. Denote by $L$ the intersection point of the segment $K O$ with the circle $s$. We want to prove that $K L > K P_{i}$. It is enough to prove that $\angle K P_{i} L > \angle K L P_{i}$. However, it is obvious that $\angle L P_{i} O \leq 30^{\circ}$ and $\angle L O P_{i} \leq 30^{\circ}$, hence $\angle K L P_{i} \leq 60^{\circ}$, while $\angle N P_{i} L = 90^{\circ} - \angle L P_{i} O \geq 60^{\circ}$. This implies that $\angle K P_{i} L \geq \angle N P_{i} L \geq 60^{\circ} \geq \angle K L P_{i}$ ( $N$ is the point on the edge of $T_{i}$ as shown in the figure). Our claim is thus proved. Now we see that the number of disks with centers in $T_{i}$ that intersect $S$ is less than or equal to 2003, and the total number of disks that intersect $S$ is not greater than $2002 + 6 \cdot 2003 = 7 \cdot 2003 - 1$.