Divide into two parts, each part containing an odd number of edges of $P$, then this diagonal is called a "good edge". It is stipulated that each edge of $P$ is a "good edge".

Given 2003 non-intersecting diagonals inside $P$ that partition $P$ into several triangles. How many isosceles triangles with two "good edges" can there be at most under this partition?
If an isosceles triangle has two good sides, it is simply called a "good triangle".
Let $A B C$ be a "good triangle", and $A B, B C$ be the good sides. Then, between $A$ and $B$, there are an odd number of edges, and the same applies between $B$ and $C$. We say these edges belong to the "good triangle" $A B C$.

In each of these groups, at least one edge does not belong to any other "good triangle". This is because two vertices between $A$ and $B$ in a "good triangle" have two equal-length sides, thus, there are an even number of edges belonging to it. Removing all edges that belong to any other "good triangle", there must be one edge left that does not belong to any other "good triangle". We designate these two edges (one in each group) to belong to triangle $A B C$.

For each "good triangle", we designate a pair of edges, with no two triangles sharing the designated edges. It follows that under this partition, there are at most 1003 "good triangles", and it is easy to draw a partition that achieves this value.