Let $P_1$ be a regular $n$-gon, where $n\in\mathbb{N}$. We construct $P_2$ as the regular $n$-gon whose vertices are the midpoints of the edges of $P_1$. Continuing analogously, we obtain regular $n$-gons $P_3,P_4,\ldots ,P_m$. For $m\ge n^2-n+1$, find the maximum number $k$ such that for any colouring of vertices of $P_1,\ldots ,P_m$ in $k$ colours there exists an isosceles trapezium $ABCD$ whose vertices $A,B,C,D$ have the same colour.

[i]Radu Ignat[/i]
1. **Understanding the Problem:**
   We start with a regular $n$-gon $P_1$. We construct $P_2$ as the regular $n$-gon whose vertices are the midpoints of the edges of $P_1$. This process continues to form $P_3, P_4, \ldots, P_m$. We need to find the maximum number $k$ such that for any coloring of the vertices of $P_1, \ldots, P_m$ in $k$ colors, there exists an isosceles trapezium $ABCD$ whose vertices $A, B, C, D$ have the same color.

2. **Key Observations:**
   - Each $P_i$ is a regular $n$-gon.
   - The vertices of $P_2$ are the midpoints of the edges of $P_1$, and this pattern continues.
   - For $m \ge n^2 - n + 1$, we need to find the maximum $k$ such that there exists an isosceles trapezium with vertices of the same color.

3. **Monochromatic Segments:**
   If we use at most $n - 1$ colors, then by the Pigeonhole Principle (PHP), there will be at least one color that appears on at least $\left\lceil \frac{n^2 - n + 1}{n - 1} \right\rceil = n$ vertices. This is because there are $n^2 - n + 1$ vertices in total, and if we distribute them among $n - 1$ colors, at least one color must appear on at least $n$ vertices.

4. **Parallel Lines and Isosceles Trapezium:**
   - The sides and diagonals of each $n$-gon are parallel to $n$ different lines.
   - These lines are the same for all $m$ polygons.
   - By PHP, at least two monochromatic segments must be parallel.
   - Since the polygons are inscribed in concentric circles, these segments cannot be diameters, thus forming an isosceles trapezium.

5. **Coloring with $n$ Colors:**
   - If we use $n$ colors, we can color the vertices such that no two segments of the same color form an isosceles trapezium.
   - This can be done by coloring all vertices on the same line $OA_k$ with the same color $C_k$, where $O$ is the common center of the polygons and $P_1 = A_1A_2 \ldots A_n$.
   - In this case, the only possibility to find four points of the same color is when they are collinear, so they cannot form a trapezoid.

6. **Conclusion:**
   - The maximum number $k$ such that for any coloring of vertices of $P_1, \ldots, P_m$ in $k$ colors there exists an isosceles trapezium $ABCD$ whose vertices $A, B, C, D$ have the same color is $n - 1$.

The final answer is $\boxed{n - 1}$