Consider $2011^2$ points arranged in the form of a $2011 \times 2011$ grid. What is the maximum number of points that can be chosen among them so that no four of them form the vertices of either an isosceles trapezium or a rectangle whose parallel sides are parallel to the grid lines?
1. **Understanding the Problem:**
   We need to find the maximum number of points that can be chosen from a \(2011 \times 2011\) grid such that no four points form the vertices of either an isosceles trapezium or a rectangle whose parallel sides are parallel to the grid lines.

2. **Initial Considerations:**
   - For an isosceles trapezium with parallel sides parallel to the grid lines, we need at least 3 points in a row or column.
   - For a rectangle with sides parallel to the grid lines, we need at least 2 points in two different rows and columns.

3. **Induction Hypothesis:**
   We will use strong induction to prove that for an \(n \times n\) grid, the maximum number of points that can be chosen without forming the specified shapes is \(3n - 3\).

4. **Base Case:**
   For \(n = 1\), the maximum number of points is \(3 \cdot 1 - 3 = 0\), which is trivially true since there are no points to choose.

5. **Inductive Step:**
   Assume the hypothesis is true for all grids of size \(k \times k\) where \(k < n\). We need to show it holds for an \(n \times n\) grid.

6. **Row and Column Analysis:**
   - Consider the row with the maximum number of points. Let this row have \(k\) points, where \(k \leq n-1\).
   - On these \(k\) columns, we can have at most \(k\) additional points to avoid forming a rectangle.

7. **Reduction to Smaller Grid:**
   - Remove the \(k\) columns and the row with \(k\) points. This leaves an \((n-1) \times (n-k)\) grid.
   - By the induction hypothesis, this smaller grid can have at most \(3(n-1) - 3\) points.

8. **Combining Points:**
   - The total number of points is the sum of points in the row with \(k\) points and the points in the remaining grid.
   - Therefore, the total number of points is \(k + (3(n-1) - 3)\).

9. **Maximizing the Points:**
   - To maximize the number of points, we need to consider the worst-case scenario where \(k = n-1\).
   - Thus, the total number of points is \((n-1) + (3(n-1) - 3) = 3n - 3\).

10. **Conclusion:**
    - By induction, the maximum number of points that can be chosen from an \(n \times n\) grid without forming the specified shapes is \(3n - 3\).

The final answer is \(\boxed{3n - 3}\).