Given a circle and $2006$ points lying on this circle. Albatross colors these $2006$ points in $17$ colors. After that, Frankinfueter joins some of the points by chords such that the endpoints of each chord have the same color and two different chords have no common points (not even a common endpoint). Hereby, Frankinfueter intends to draw as many chords as possible, while Albatross is trying to hinder him as much as he can. What is the maximal number of chords Frankinfueter will always be able to draw?
We need to prove that Frankinfueter can always draw at least 117 chords, regardless of how Albatross colors the points. We will use strong induction on \( n \) to show that for any \( c \), given \( nc \) points and \( c \) colors, we can always draw \( n-1 \) chords.

1. **Base Case: \( n = 2 \)**
   - If \( n = 2 \), we have \( 2c \) points and \( c \) colors. Since there are at least 2 points of each color, we can always draw at least 1 chord. This is trivial and serves as our base case.

2. **Inductive Step:**
   - Assume that for any \( k \leq n \), given \( kc \) points and \( c \) colors, we can always draw \( k-1 \) chords. We need to show that for \( n+1 \), given \( (n+1)c \) points and \( c \) colors, we can always draw \( n \) chords.

3. **Color Distribution:**
   - There must be at least one color that appears at least \( n+1 \) times among the \( (n+1)c \) points. Let's call this color blue.

4. **Case 1: Blue appears exactly \( n+1 \) times.**
   - **Subcase 1.1: There are 2 blue points with at most \( c-2 \) points between them.**
     - Remove these two blue points and the points between them. We are left with at least \( nc \) points and \( c \) colors. By the induction hypothesis, we can draw \( n-1 \) chords among these points. Adding the chord between the two blue points, we have drawn \( n \) chords in total.
   
   - **Subcase 1.2: For every pair of adjacent blue points, there are exactly \( c-1 \) points between them.**
     - Enumerate the blue points clockwise as \( 1, 2, \ldots, n+1 \). Draw chords between pairs \( (1, 2), (3, 4), \ldots \). This gives us \( \left\lfloor \frac{n+1}{2} \right\rfloor \) chords. The remaining points (those not between two blue endpoints of a chord) are \( \left\lceil \frac{n+1}{2} \right\rceil (c-1) \) in number. By the induction hypothesis, we can draw \( \left\lceil \frac{n+1}{2} \right\rceil - 1 \) more chords among these points. Thus, we have drawn \( \left\lfloor \frac{n+1}{2} \right\rfloor + \left\lceil \frac{n+1}{2} \right\rceil - 1 = n \) chords in total.

5. **Case 2: Blue appears at least \( n+2 \) times.**
   - There are at least two blue points with at most \( c-2 \) points between them. Remove these two blue points and the points between them. We are left with at least \( nc \) points and \( c \) colors. By the induction hypothesis, we can draw \( n-1 \) chords among these points. Adding the chord between the two blue points, we have drawn \( n \) chords in total.

By induction, we have shown that for any \( n \), given \( nc \) points and \( c \) colors, we can always draw \( n-1 \) chords. Specifically, for \( n = 118 \) and \( c = 17 \), we have \( 2006 \) points and can always draw \( 117 \) chords.

The final answer is \( \boxed{ 117 } \).