23. Find the smallest positive integer $n(\geqslant 3)$, such that in any set of $n$ points in the plane with no three points collinear, there must be three points that are the vertices of a non-isosceles triangle.
(2005 China National Training Team Problem)
23. First, when 6 points in a plane are the 5 vertices of a regular pentagon and its center, all triangles formed by these 6 points are isosceles triangles, so the required positive integer \( n \geqslant 7 \). Next, 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 \( E \). Without loss of generality, 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 \( l_{1} = \overparen{M A N} \cup \overparen{M B N} \), and \( l_{2} = \) 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 three points in \( E \) are collinear, there are at most 2 points of \( E \) on \( l_{2} \), so at least \( 7 - 2 - 2 = 3 \) points of \( E \) are on \( \overparen{A M} \), \( \overparen{B M} \), \( \overparen{A N} \), \( \overparen{B N} \) (excluding endpoints). Without loss of generality, assume there is a point \( C \in E \) on \( \overparen{B N} \) (where \( C \neq B \), \( C \neq N \)), and \( \overparen{B C} \) contains no points of \( E \). Draw arcs centered at \( A \) and \( C \) with radius \( A C = A B \) intersecting at points \( M' \) and \( N' \). Let \( l_{3} = \overparen{M' A N'} \cup \overparen{M' C N'} \), and \( l_{4} = \) line segment \( M' N' \). Similarly, all points of \( E \) belong to \( l_{3} \cup l_{4} \). Therefore, \( E \subseteq (l_{1} \cup l_{2}) \cap (l_{3} \cup l_{4}) \), and \( \overparen{B C} \) contains no points of \( E \). So, \( E \subseteq \overparen{M' B} \cup \overparen{C N} \cup \{A\} \cup \{O\} \cup \{T\} \cup \{R\} \), where \( O \) is the intersection of \( l_{2} \) and \( l_{4} \), \( T \) is the intersection of \( l_{2} \) and \( \overparen{M' A} \), and \( R \) is the intersection of \( l_{4} \) and \( \overparen{A N} \). Let \( S_{1} = \overparen{M' B} \cup \{T\} \), \( S_{2} = \overparen{C N} \cup \{R\} \), then \( S_{1} \cup S_{2} \) contains at least \( 7 - 3 - 1 = 3 \) points of \( E \) different from \( A \), \( B \), and \( C \). So, \( S_{1} \) or \( S_{2} \) contains at least 2 points of \( E \) different from \( A \), \( B \), and \( C \). Without loss of generality, assume \( S_{2} \) contains 2 points \( P \) and \( Q \) of \( E \) different from \( A \), \( B \), and \( C \), then there are only two cases: (1) \( P \in \overparen{C N} \), \( Q \in \overparen{C N} \). 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 \( Q C = B C \). So, \( C P = C Q \), i.e., \( C \) lies on the perpendicular bisector of \( P Q \), but the perpendicular bisector of \( P Q \) only passes through \( A \) and not through \( C \), a contradiction. (2) \( P \in \overparen{C N} \), \( Q = R \). Similarly, we get \( P C = B C \). Since \( \triangle R B C \) is an isosceles triangle and \( B R = B N > B C \), we have \( R C = B C \), so \( R C = P C = B C \). 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 summary, the smallest positive integer \( n = 7 \).