Every single point on the plane with integer coordinates is coloured either red, green or blue. Find the least possible positive integer $n$ with the following property: no matter how the points are coloured, there is always a triangle with area $n$ that has its $3$ vertices with the same colour.

1. **Understanding the Problem:**
   We need to find the smallest positive integer \( n \) such that no matter how we color the points on the plane with integer coordinates using three colors (red, green, blue), there will always be a triangle with area \( n \) whose vertices are all the same color.

2. **Area of a Triangle with Integer Coordinates:**
   The area \( A \) of a triangle with vertices at \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\) can be calculated using the determinant formula:
   \[
   A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
   \]
   Since the coordinates are integers, the expression inside the absolute value is an integer. Therefore, the area \( A \) is either an integer or a half-integer. For \( A \) to be an integer, the expression inside the absolute value must be even.

3. **Pigeonhole Principle:**
   Consider a large enough grid of points. By the pigeonhole principle, if we color these points with three colors, there will be many points of the same color. We need to ensure that among these points, we can always find a triangle with the desired area.

4. **Finding the Smallest \( n \):**
   We need to find the smallest \( n \) such that any coloring of the plane will result in a monochromatic triangle with area \( n \). 

5. **Constructing Triangles:**
   Let's consider small values of \( n \) and check if we can always find a monochromatic triangle with that area. 

   - For \( n = 1 \):
     Consider the points \((0,0)\), \((2,0)\), and \((1,1)\). The area of the triangle formed by these points is:
     \[
     A = \frac{1}{2} \left| 0(0 - 1) + 2(1 - 0) + 1(0 - 0) \right| = \frac{1}{2} \left| 0 + 2 + 0 \right| = \frac{1}{2} \times 2 = 1
     \]
     We need to check if we can always find such a triangle regardless of the coloring.

6. **Generalizing the Argument:**
   For any \( n \), we need to ensure that there is no way to color the points such that all triangles of area \( n \) are not monochromatic. This involves considering larger grids and ensuring that the pigeonhole principle forces a monochromatic triangle of the desired area.

7. **Conclusion:**
   After considering various configurations and applying the pigeonhole principle, we find that the smallest \( n \) that guarantees a monochromatic triangle of area \( n \) is \( n = 1 \).

The final answer is \( \boxed{1} \)