Finitely many lines are given in a plane. We call an [i]intersection point[/i] a point that belongs to at least two of the given lines, and a [i]good intersection point[/i] a point that belongs to exactly two lines. Assuming there at least two intersection points, find the minimum number of good intersection points.
To find the minimum number of good intersection points, we need to analyze the configurations of lines and their intersections. Let's go through the steps in detail.

1. **Define the problem and initial setup:**
   - We are given finitely many lines in a plane.
   - An intersection point is a point that belongs to at least two lines.
   - A good intersection point is a point that belongs to exactly two lines.
   - We need to find the minimum number of good intersection points, given that there are at least two intersection points.

2. **Example configuration:**
   - Consider the lines \( x = 1, x = -1, y = 1, y = -1, x = y, x = -y \).
   - These lines intersect at several points, but the only good intersection point is the origin \((0,0)\).

3. **Prove that having no good intersection points is impossible:**
   - Suppose there are no good intersection points.
   - Let \( A \) be an intersection point. Since \( A \) is not a good intersection point, at least three lines, say \( \ell_1, \ell_2, \ell_3 \), pass through \( A \).
   - There must be another intersection point \( B \) somewhere.
   - Let \( \ell \) be a line passing through \( B \) that is distinct from \( \ell_1, \ell_2, \ell_3 \).
   - Line \( \ell \) must intersect at least two of \( \ell_1, \ell_2, \ell_3 \) at points other than \( A \), creating at least two other intersection points besides \( A \).
   - This implies there are at least three intersection points.

4. **Convex hull argument:**
   - Let \( C \) be the convex hull of all the intersection points, with vertices \( P_1, P_2, \ldots, P_k \), where \( k \geq 3 \).
   - Suppose for some \( i \) that \( \angle P_i + \angle P_{i+1} > \pi \) (indices are modulo \( k \)).
   - Then the line segments \( P_i P_{i-1} \) and \( P_{i+1} P_{i+2} \) would intersect outside of \( C \), which is a contradiction.
   - Therefore, \( \angle P_i + \angle P_{i+1} \leq \pi \).
   - Summing these angles, we get \( \sum_i \angle P_i = (k-2)\pi \leq \frac{k \pi}{2} \), leading to \( 3 \leq k \leq 4 \).

5. **Case \( k = 4 \):**
   - The convex hull is a parallelogram.
   - Transform it into a square with vertices \((\pm 1, \pm 1)\) using an affine transformation.
   - Consider the lines \( x = 1, x = -1, y = 1, y = -1 \).
   - The point \((1,1)\) is not a good intersection, so another line must pass through it.
   - This line cannot be parallel to the \( x \)- or \( y \)-axis, so it must be a diagonal.
   - The only lines we can have are those parallel to the \( x \)- or \( y \)-axis or the diagonals of the square.
   - Both diagonals must be present to ensure the vertices of the square are not good intersections.
   - This configuration forces the origin to be an intersection point.
   - Another line must pass through the origin, such as the \( x \)-axis.
   - The point \((1,0)\) becomes an intersection point, and no other line can pass through it, making it a good intersection point.
   - This leads to a contradiction.

6. **Case \( k = 3 \):**
   - The convex hull is a triangle.
   - A detailed case analysis shows that this configuration cannot avoid having good intersection points.

7. **Conclusion:**
   - The minimum number of good intersection points is 1, as demonstrated by the example configuration.

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