28.2.9 * Find the smallest positive integer $n(n \geqslant 3)$, such that in any set of $n$ points in the plane with no three points collinear, there are two points that are vertices of a non-isosceles triangle.
First, if 6 points in a plane are the 5 vertices and the center of a regular pentagon, all triangles formed by these 6 points are isosceles triangles. Therefore, the required positive integer \( n \geqslant 7 \).

Second, if there exist 7 points in a plane (where no three points are collinear), such that all triangles formed by these 7 points are isosceles triangles. Let the set of these 7 points be \( E \).
Assume the two farthest points in \( E \) are \( A \) and \( B \). Draw arcs centered at \( A \) and \( B \) with radius \( AB \), intersecting at points \( M \) and \( N \).
Let \( l_{1} = \overparen{M A N} \cup \sqrt{M B N} \), \( l_{2} = \) line segment \( M N \).
Take any \( P \in E, P \neq A, P \neq B \), then \( \triangle P A B \) is an isosceles triangle. Thus, at least one of the following three cases holds: (1) \( P A = P B \leqslant A B \); (2) \( P A = A B \geqslant P B \); (3) \( P B = A B \geqslant P A \), so \( P \in l_{1} \cup l_{2} \).
Since no two points in \( E \) are collinear, at most 2 points of \( E \) can lie on \( l_{2} \), thus at least \( 7 - 2 - 2 = 3 \) points of \( E \) lie on \( \overparen{A M} \), \( \overparen{B M} \), \( \overparen{A N} \), \( \overparen{B N} \) (excluding endpoints).
Assume \( \overparen{B N} \) has one point \( C \) of \( E \) (where \( C \neq B, C \neq N \)), and there are no points of \( E \) inside \( B C \). Draw arcs centered at \( A \) and \( C \) with radius \( A C = A B \), intersecting at points \( M' \) and \( N' \).
Let \( l_{3} = \overparen{M' A N'} \cup \widetilde{M' C N} \), \( l_{4} = \) line segment \( M' N' \). Similarly, all points of \( E \) belong to \( l_{3} \cup l_{4} \). So \( \{O\} \cup \{T\} \cup \{R\} \), where \( O \) is the intersection of \( l_{2} \) and \( l_{4} \), \( T \) is the intersection of \( l_{2} \) and \( \overparen{M A} \), \( R \) is the intersection of \( l_{1} \) and \( \overparen{A N} \).
Let \( S_{1} = \overparen{M' B} \cup \{T\} \), \( S_{2} = \overparen{C N} \cup \{R\} \), then \( S_{1} \cup S_{2} \) has at least \( 7 - 3 - 1 = 3 \) points of \( E \) different from \( A \), \( B \), and \( C \). Therefore, \( S_{1} \) or \( S_{2} \) has at least 2 points of \( E \) different from \( A \), \( B \), and \( C \). Assume \( S_{2} \) has 2 points \( P \) and \( Q \) of \( E \) different from \( A \), \( B \), and \( C \), then there are only the following two cases.
(1) \( P \in \overparen{C N} \), \( Q \in \overparen{C N} \) (as shown in Figure a). Since \( \triangle P B C \) is an isosceles triangle and \( P B > B C \), we have \( P C = B C \). Similarly, from \( \triangle Q B C \) being an isosceles triangle, we get \( Q C = B C \). Therefore, \( C P = C Q \), meaning \( C \) lies on the perpendicular bisector of \( P Q \), but the perpendicular bisector of \( P Q \) only passes through \( A \) and not \( C \), which is a contradiction.
(2) \( P \in \overparen{C N} \), \( Q = R \) (as shown in Figure b), similarly we get \( P C = B C \).
Also, \( \triangle R B C \) is an isosceles triangle, and \( B R = B N > B C \), so \( R C = B C \), thus \( R C = P C = B C \). In \( \triangle B C P \) and \( \triangle B C R \), \( B C = B C \), \( P C = R C \), \( \angle B C P \geqslant \angle B C N > \angle B C R \), so \( B P > B R \). This contradicts \( B P \leqslant B N = B R \).
Therefore, there do not exist 7 points in a plane (where no three points are collinear) such that all triangles formed by these 7 points are isosceles triangles.
In conclusion, the smallest positive integer \( n = 7 \).