15 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 must be three points that are the vertices of a non-isosceles triangle.
Solution: First, if the 6 points in the plane are the 5 vertices of a regular pentagon and its center, 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 the 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 with $A$ and $B$ as centers and $AB$ as the radius, intersecting at points $M$ and $N$.
Let $l_{1}=\widehat{MAN} \cup \widehat{MBN}$, $l_{2}=$ line segment $MN$.
Take any $P \in E, P \neq A, P \neq B$, then $\triangle PAB$ is an isosceles triangle. Thus, at least one of the following three cases must hold: (1) $PA=PB \leqslant AB$; (2) $PA=AB \geqslant PB$; (3) $PB=AB \geqslant PA$, 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 $\overparen{AM}, \overparen{BM}, \overparen{AN}, \overparen{BN}$ (excluding endpoints) contain at least $7-2-2=3$ points of $E$.

Without loss of generality, assume $\overparen{BN}$ has one point $C (C \neq B, C \neq N)$, and $\overparen{BC}$ contains no points of $E$. Draw arcs with $A$ and $C$ as centers and $AC=AB$ as the radius, intersecting at points $M'$ and $N'$.

Let $l_{3}=\widehat{M'AN'} \cup \widehat{M'CN'}$, $l_{4}=$ 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 $\overparen{BC}$ contains no points of $E$. So $E \subseteq \overparen{M'B} \cup \overparen{CN} \cup \{A\} \cup \{O\} \cup \{T\} \cup \{R\}$, where point $O$ is the intersection of $l_{2}$ and $l_{4}$, $T$ is the intersection of $l_{2}$ and $\mathscr{M}'A$, and $R$ is the intersection of $l_{4}$ and $\overparen{AN}$.
Let $S_{1}=\overparen{M'B} \cup \{T\}$, $S_{2}=\overparen{CN} \cup \{R\}$, then
$S_{1} \cup S_{2}$ contains at least $7-3-1=3$ points of $E$ different from $A, B, C$. So $S_{1}$ or $S_{2}$ contains at least 2 points of $E$ different from $A, B, C$. Without loss of generality, assume $S_{2}$ contains two points $P, Q$ of $E$ different from $A, B, C$, then there are only the following two cases.
(1) $P \in \overparen{CN}, Q \in \overparen{CN}$. Since $\triangle PBC$ is an isosceles triangle, and $PB > BC$, so $PC = BC$. Similarly, from $\triangle QBC$ being an isosceles triangle, we get $QC = BC$. Therefore, $CP = CQ$, i.e., $C$ lies on the perpendicular bisector of $PQ$, but the perpendicular bisector of $PQ$ only passes through $A$ and not through $C$, a contradiction.
(2) $P \in \overparen{CN}, Q = R$, similarly, we get $PC = BC$.
Also, $\triangle RBC$ is an isosceles triangle, and $BR = BN > BC$, so $RC = BC$, so $RC = PC = BC$. In $\triangle BCP$ and $\triangle BCR$, $BC = BC, PC = RC, \angle BCP \geqslant \angle BCN > \angle BCR$, so $BP > BR$. This contradicts $BP \leqslant BN = BR$.

It is evident that there do not exist 7 points in the 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 $n=7$.