Solution:
We give a proof by contradiction. In addition to the hypothesis, we assume that a disk (without loss of generality let this be $K_{1}$) intersects at least $14021$ other disks. The centers of these disks then obviously all lie in a disk of radius $2r$ about the center $M_{1}$ of $K_{1}$. If $2003$ or more of the centers of these other disks were to lie in or on the boundary of $K_{1}$, then all these disks would contain $M_{1}$. Then $M_{1}$ would be contained in more than $2003$ disks, since $M_{1}$ is also contained in $K_{1}$. Hence at least $14021 - 2002 = 12019$ centers of the other disks must lie in an annulus with inner radius $r$ and outer radius $2r$ about $M_{1}$. If we divide this annulus into $6$ equally sized sectors with interior angle $60^{\circ}$, then by the pigeonhole principle at least one sector contains at least $2004$ of these centers.
In the diagram this sector is shown with the midpoint $M$ of the segment $AB$. From $|AM_{1}| = |BM_{1}| = 2r$ and $\angle AM_{1}B = 60^{\circ}$ it follows that $\triangle ABM_{1}$ is equilateral. Hence $MD$ and $MC$ are its midlines, and it follows that $|MA| = |MB| = |MC| = |MD| = r$. Therefore the sector lies entirely within a circle about $M$ with radius $r$. Each of the at least $2004$ disks with centers in this sector therefore contains the point $M$. Thus $M$ is contained in

at least $2004$ disks — a contradiction to the assumption.