It is considered a regular polygon with $n$ sides, where $n(n>3)$ is an odd number that does not divide by 3. From the vertices of the polygon are arbitrarily chosen $m(0\leq m\leq n)$ vertices that are colored in red and the others in black. A triangle with the vertices at the vertices of the polygon it is considered $monocolor$ ,if all of its vertices are of the same color. Prove that the number of all $monocolor$  isosceles triangles  with the vertices at the given polygon ends does not depend on the way of coloring of the vertices of the polygon. Determine the number of all these $monocolor$ isosceles triangles.

1. **Setup and Definitions**:
   - Consider a regular polygon with \( n \) sides, where \( n \) is an odd number greater than 3 and not divisible by 3.
   - Vertices are colored either red or black, with \( m \) vertices colored red and \( n-m \) vertices colored black.
   - We need to prove that the number of monocolor isosceles triangles does not depend on the coloring and determine this number.

2. **Labeling and Families of Line Segments**:
   - Number the vertices from \( 0 \) to \( 2k \) in a clockwise manner, where \( n = 2k + 1 \).
   - Define two families of line segments:
     - **Swirls**: Line segments of the form \((x, 2x)\).
     - **Parallels**: Line segments of the form \((y, 2k+1-y)\).

3. **Potential Status of Line Segments**:
   - A line segment is **potential** if both endpoints are red.
   - A line segment is **impotential** if both endpoints are black.
   - A line segment is a **waste** if one endpoint is red and the other is black.

4. **Invariance Under Swapping**:
   - Consider changing a black vertex at \( 0 \) to red. The number of black monocolor isosceles triangles lost equals the number of impotential line segments, and the number of red monocolor isosceles triangles gained equals the number of potential line segments.
   - The difference between the number of potential and impotential line segments remains constant regardless of the arrangement of the other \( 2k \) points.

5. **Swapping Points**:
   - Swapping two points of different colors affects both swirls and parallels:
     - **Case 1**: One swirl is potential and the other is impotential. Both become waste.
     - **Case 2**: Both swirls are waste. One becomes potential, and the other becomes impotential.
     - **Case 3**: One swirl is waste, and the other is potential/impotential. Their potential status swaps.
   - The same applies to parallels.

6. **Conclusion from Swapping**:
   - The number of monocolor isosceles triangles is invariant under swapping points.
   - By repeatedly swapping points, any configuration can be transformed into any other configuration with the same number of red and black points.

7. **Counting Monocolor Isosceles Triangles**:
   - Consider \( m \) consecutive red points and \( n-m \) consecutive black points.
   - For each vertex \( i \), there are two isosceles triangles with vertex \( i \) as the apex: \((i, i+k, i-k)\) and \((i, i-k, i+k)\).
   - Since \( n \) is odd, each vertex is the apex of exactly two isosceles triangles.
   - Each triangle is counted three times (once for each vertex), so the total number of isosceles triangles is \(\frac{2n}{3}\).

8. **Monocolor Isosceles Triangles**:
   - Since the number of monocolor isosceles triangles is invariant under the coloring, we can calculate it for any specific configuration.
   - For \( m \) red vertices and \( n-m \) black vertices, the number of monocolor isosceles triangles is \(\frac{2m}{3}\) red and \(\frac{2(n-m)}{3}\) black.

The final answer is \( \boxed{ \frac{2m}{3} + \frac{2(n-m)}{3} = \frac{2n}{3} } \).