24. Let $P$ be a regular 2006-gon. If an end of a diagonal of $P$ divides the boundary of $P$ into two parts, each containing an odd number of sides of $P$, then the diagonal is called a "good diagonal". It is stipulated that each side of $P$ is also a "good diagonal". Given that 2003 non-intersecting diagonals inside $P$ divide $P$ into several triangles. How many isosceles triangles with two "good diagonals" can there be at most in such a division?
（47th IMO Problem）
24. If an isosceles triangle has two good sides, it is briefly referred to as a "good triangle". Let $\triangle A B C$ be a "good triangle", and $A B$ and $B C$ be good sides. Then, there are an odd number of sides between $A$ and $B$, and the same applies to $B$ and $C$. We say these sides belong to the "good triangle" $A B C$. In each of these groups, at least one side does not belong to any other "good triangle", because the "good triangles" with vertices between $A$ and $B$ have two equal sides, thus an even number of sides belong to it. After removing all sides that belong to any other "good triangle", there must be one side left that does not belong to any other "good triangle". We designate these two sides (one side from each group) to belong to triangle $A B C$. For each "good triangle", we designate a pair of sides, and no two "good triangles" share the designated sides. It follows that in this partition, there are at most $\frac{1}{2} \times 2006 = 1003$ "good triangles". Moreover, it is easy to draw a partition that achieves this value: consider a regular 2006-sided polygon $A_{1} A_{2} \cdots A_{2006}$, and connect 2003 diagonals $A_{2 i-1} A_{2 i+1}\left(i=1,2, \cdots, 1003\right.$, where $\left.A_{2007}=A_{1}\right)$ and $A_{1} A_{2 i+3}(i=1,2, \cdots, 1000)$. This results in exactly 1003 "good triangles": $\triangle A_{2 i-1} A_{2 i} A_{2 i+1}\left(i=1,2, \cdots, 1003\right.$, where $\left.A_{2007}=A_{1}\right)$. Therefore, there are at most 1003 isosceles triangles with two sides being "good sides".