Example 10 Suppose there are 2009 points in a plane, among any 17 points, there exist 11 points that can be covered by a circle of diameter 1. If $n$ circles of diameter 2 can cover all 2009 points. Find the minimum value of $n$.
First, construct a regular hexagon $A B C D E F$ with a side length of 3, and let the center of the hexagon be $O$. Choose a set of 2009 points $S$ as follows: 6 points are the 6 vertices of the hexagon, and the remaining 2003 points are all within a circle centered at $O$ with a diameter of 1. Clearly, in any 17 points of $S$, at least 11 points can be covered by a circle centered at $O$ with a diameter of 1. Moreover, any circle with a diameter of 2 can cover at most one vertex of the hexagon and cannot cover any other points in $S$, so at least 7 circles with a diameter of 2 are needed to cover all 2009 points in $S$, hence $n \geqslant 7$.
Next, we prove that at most 7 circles with a diameter of 2 are sufficient to cover all 2009 points in $S$.
For any 8 points in $S$, take another 9 points from $S$, making a total of 17 points. By the given condition, at least 11 of these 17 points can be covered by a circle with a diameter of 1, leaving at most 6 points uncovered. Therefore, among the 8 points chosen, at least 2 points can be covered by a circle with a diameter of 1, meaning the distance between these 2 points is no more than 1. Thus, we have proven that in any 8 points of $S$, there must be 2 points whose distance is no more than 1.

Take any point $A_{1}$ in $S$, and draw a circle $\odot A_{1}$ with $A_{1}$ as the center and a diameter of 2. If all other points in $S$ are covered by this circle, then only 1 circle is needed. Otherwise, take any point $A_{2}$ outside $\odot A_{1}$, then $A_{1} A_{2}>1$. Draw a circle $\odot A_{2}$ with $A_{2}$ as the center and a diameter of 2. If all points in $S$ are covered by $\odot A_{1}$ and $\odot A_{2}$, then only 2 circles are needed. Otherwise, continue this process to obtain 7 circles: $\odot A_{1}, \odot A_{2}, \cdots, \odot A_{7}$, each with a diameter of 2 and $A_{i} A_{j}>1(1 \leqslant i<j \leqslant 7)$. Therefore, there must be some $A_{i}$ such that $A_{i} B \leqslant 1$, meaning $B$ is covered by $\odot A_{i}$. By the arbitrariness of $B$, all points in $S$ are covered by the 7 circles: $\odot A_{1}, \odot A_{2}, \cdots, \odot A_{7}$.
In conclusion, the minimum value of $n$ is 7.