Among the coordinates $(x,y)$ $(1\leq x,y\leq 101)$, choose some points such that there does not exist $4$ points which form a isoceles trapezium with its base parallel to either the $x$ or $y$ axis(including rectangles). Find the maximum number of coordinate points that can be chosen.
1. **Define the problem and constraints:**
   We need to choose points \((x, y)\) where \(1 \leq x, y \leq 101\) such that no four points form an isosceles trapezium with its base parallel to either the \(x\)-axis or \(y\)-axis. This includes avoiding rectangles.

2. **Identify dangerous rows:**
   A row is considered dangerous if it contains at least 2 of the chosen points. If no row is dangerous, then each row can have at most 1 point, leading to a maximum of 101 points.

3. **Consider the case with one dangerous row:**
   If there is one dangerous row, it can have at most 2 points to avoid forming an isosceles trapezium with another row. This means the dangerous row can contribute 2 points.

4. **Calculate the total points:**
   - The dangerous row contributes 2 points.
   - The remaining 100 rows can each contribute 1 point.

   Therefore, the total number of points is:
   \[
   1 \times 100 + 101 = 201
   \]

5. **Verify the solution:**
   - If there are more than 2 points in any row, it would be possible to form an isosceles trapezium with another row.
   - If there are more than 1 dangerous row, it would also be possible to form an isosceles trapezium.

Thus, the maximum number of points that can be chosen without forming an isosceles trapezium is 201.

The final answer is \(\boxed{201}\).