Let $n\ge 3$ be a positive integer. In the plane $n$ points which are not all collinear are marked. Find the least possible number of triangles whose vertices are all marked. 

(Recall that the vertices of a triangle are not collinear.)

1. **Understanding the Problem:**
   We need to find the least possible number of triangles that can be formed by \( n \) points in the plane, where not all points are collinear. Recall that a triangle is formed by three non-collinear points.

2. **Initial Consideration:**
   If all \( n \) points were collinear, no triangles could be formed. However, since not all points are collinear, at least one point must be off the line formed by any subset of collinear points.

3. **Optimal Configuration:**
   To minimize the number of triangles, we should maximize the number of collinear points. Suppose \( n-1 \) points are collinear, and one point is not on this line. This configuration will help us achieve the minimum number of triangles.

4. **Counting Triangles:**
   - If \( n-1 \) points are collinear, any triangle must include the one point that is not collinear with the others.
   - The number of ways to choose 2 points from the \( n-1 \) collinear points is \( \binom{n-1}{2} \).
   - Each pair of collinear points, combined with the one non-collinear point, forms a unique triangle.

5. **Verification:**
   - The number of triangles formed in this configuration is exactly \( \binom{n-1}{2} \).
   - This is because each triangle must include the non-collinear point, and the other two vertices must be chosen from the \( n-1 \) collinear points.

6. **General Case:**
   - If we add a new point that is not on any existing line, the number of triangles increases by \( \binom{n}{2} \).
   - If the new point is on a line with \( k \) points, the number of new triangles formed is \( \binom{n+1-k}{2} + (k-1)(n+1-k) \).

7. **Minimizing the Number of Triangles:**
   - The expression \( \binom{n+1-k}{2} + (k-1)(n+1-k) \) simplifies to \( \frac{(n+1-k)(n+k-2)}{2} \).
   - The term \( 3k - k^2 \) attains its minimum value when \( k \) is as large as possible, i.e., \( k = n \).
   - When \( k = n \), the increase in the number of triangles is \( n-1 \), which is less than \( \binom{n}{2} \).

8. **Conclusion:**
   - The least number of triangles is achieved when \( n-1 \) points are collinear, and one point is not collinear with them.
   - The number of triangles in this configuration is \( \binom{n-1}{2} \).

\[
\boxed{\binom{n-1}{2}}
\]