(9. 3 There are 2006 points on a circle. Petya first colors these points using 17 colors, then Kolya draws chords between some of these 2006 points, such that the endpoints of each chord are the same color, and no two chords intersect (including at endpoints). Kolya wants to draw as many such chords as possible, while Petya tries to hinder Kolya. Question: What is the maximum number of chords Kolya can draw?
Answer: 117.
Notice that $2006=17 \cdot 118$; hence there exist two colors such that the total number of points of these two colors is no less than $2 \cdot 118=236$.

We will prove by induction on $k$ that for at most 2 colors of $2 k-1$ points on a circle, it is always possible to find $k-1$ non-intersecting monochromatic chords. For $k=2$, this is obvious. Assume $k>2$. Choose two points of the same color such that the arc between them contains no other points. Connect them to form a monochromatic chord, and apply the induction hypothesis to the remaining points.

Now, take at most 235 points of at most 2 colors, and apply the above conclusion to find that Kolya can always find 117 monochromatic chords. Below is an example showing that more than 117 monochromatic chords cannot be found.

Suppose there are $17 k$ points on the circle, sequentially labeled as $1,2, \cdots, 17 k$. Assume Petya colors them with 17 colors, 0, 1, 2, ..., 16, based on their remainders when divided by 17. We will prove by induction that it is impossible to draw more than $k-1$ non-intersecting monochromatic chords. For $k=1$, the conclusion is obvious. Now assume the conclusion holds for $<k$. Suppose there are $17 k$ points on the circle, colored according to the above rule. Let $AB$ be the shortest monochromatic chord. Then there are no other monochromatic chords on one side of this chord, and on the other side, there are $17 l, l<k$ points. By the induction hypothesis, there are at most $l-1$ monochromatic chords on this side. Thus, there are a total of $l-1+1<k$ monochromatic chords.