Four. (50 points) Divide the sides of the equilateral $\triangle ABC$ into four equal parts, and draw lines parallel to the other two sides through each division point. The 15 points formed by the intersections of the sides of $\triangle ABC$ and these parallel lines are called grid points. Among these 15 grid points, if $n$ points are chosen, there will definitely be three points that can form an isosceles triangle (including equilateral triangles). Find the minimum value of $n$.

In the 15 grid points, if $n$ points are chosen, there will definitely be three points that can form an isosceles triangle (including equilateral triangles). Find the minimum value of $n$.
Four, Solution: The minimum value of $n$ is 6. Let the three equal division points from $A$ to $B$ on side $AB$ be $L, F, W$; the three equal division points from $B$ to $C$ on side $BC$ be $Y, D, U$; and the three equal division points from $C$ to $A$ on side $CA$ be $N, E, M$. The intersection points of $MV, LU, EF$ are $P$; the intersection points of $MV, WN, DF$ are $Q$; and the intersection points of $WN, LU, DE$ are $R$. If the minimum value of $n$ is 5, and 5 grid points are taken on side $AB$, then there do not exist three grid points that can form an isosceles triangle, hence $n \geqslant 6$.

Below, we prove that if $n$ grid points are taken from these 15 grid points such that no three points can form an isosceles triangle, then $n \leqslant 5$.

Assume the $n$ points taken are red points, and the other points are blue points. Since $P, Q, R$ cannot all be red points, we discuss the following three cases.
(1) If only two of $P, Q, R$ are red points, assume $Q, R$ are red points. Thus, $P, A, D$ are all blue points, and $U, V$ and $E, F$ are also all blue points. Since at most one of $M, C$ can be a red point (otherwise $M, C$ and $R$ form an isosceles triangle), at most one of $N, C$ can be a red point (otherwise $N, C$ and $R$ form an isosceles triangle), and at most one of $M, N$ can be a red point (otherwise $M, N$ and $Q$ form an isosceles triangle), at most one of $M, N, C$ can be a red point. Similarly, at most one of $L, W, B$ can be a red point, hence $n \leqslant 5$.
(2) If only one of $P, Q, R$ is a red point, assume $P$ is the red point.
(1) If $D$ is a red point, then $W, N$ are both blue points, and at most one of $F, L, M, E$ can be a red point. If $L$ (or $M$) is a red point, then $A$ is a blue point, and at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point, hence $n \leqslant 5$. If $F$ (or $E$) is a red point, then $A$ is a blue point, and at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point, hence $n \leqslant 5$.
(2) If $D$ is a blue point and $A$ is a red point, then $L, M$ and $W, N$ are all blue points, and at most one of $E, F$ can be a red point; at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point, hence $n \leqslant 5$.
(3) If $D, A$ are both blue points, since at most one of $L, M$ can be a red point, assume $L$ is a red point, then $E, F, M$ are all blue points, and at most one of $W, N$ can be a red point; at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point, hence $n \leqslant 5$.
(4) If $D, A, L, M$ are all blue points, when $E, F$ are both red points, then $W, N$ and $B, C$ are all blue points, and at most one of $U, V$ can be a red point, hence $n \leqslant 5$. When at most one of $E, F$ is a red point, at most one of $W, N$ can be a red point; at most one of $B, C$ can be a red point; at most one of $U, V$ can be a red point, hence $n \leqslant 5$.
(3) If $P, Q, R$ are all blue points, when each of the three-point groups $A, L, M; B, V, W; C, N, U; D, E, F$ has at most one red point, then there are at most four red points, otherwise, in $A, L, M; B, V, W; C, N, U; D, E, F$, there must be a three-point group that contains at least two red points.
(1) If $D, E, F$ contain at least two red points, assume $E, F$ are red points, then $D, A$ and $B, C$ are all blue points, and at most one of $L, M$ can be a red point; at most one of $W, V$ can be a red point; at most one of $N, U$ can be a red point, hence $n \leqslant 5$.
(2) If $A, L, M; B, V, W; C, N, U$ contain a three-point group with at least two red points, assume $B, V, W$ contain at least two red points.

If $V, W$ are red points, then $B, E$ and $F, D$ are all blue points. If one of $L, U$ is a red point, assume $L$ is a red point, then $C, N$ are all blue points, and at most one of $A, M$ can be a red point, even if $U$ is a red point, $n \leqslant 5$. If $L, U$ are both blue points, then at most one of $M, C$ can be a red point; at most one of $A, N$ can be a red point, hence $n \leqslant 5$.
If $B, V$ are red points, then $W$ is a blue point.
(1) If $F$ is a red point, then $D, M, N$ are all blue points. If $E$ is a red point, then $C, A$ are all blue points, and at most one of $L, U$ can be a red point, hence $n \leqslant 5$. If $E$ is a blue point, at most one of $A, C$ can be a red point, and at most one of $L, U$ can be a red point, hence $n \leqslant 5$.
(ii) If $F$ is a blue point and $D$ is a red point, then $E, L$ are both blue points, and at most one of $A, C$ can be a red point; at most one of $M, N, U$ can be a red point, hence $n \leqslant 5$.
(iii) If $F, D$ are blue points and $L$ is a red point, then $U, N$ are both blue points, and at most one of $A, C$ can be a red point; at most one of $M, E$ can be a red point, hence $n \leqslant 5$.
(iv) If $F, D, L$ are blue points and $U$ is a red point, then $A$ is a blue point, and at most one of $M, C$ can be a red point; at most one of $N, E$ can be a red point, hence $n \leqslant 5$.
(V) If $F, D, L, U$ are blue points, then at most one of $M, N$ can be a red point; at most one of $A, C$ can be a red point, even if $E$ is a red point, $n \leqslant 5$.
Using the contrapositive, we get that when $n \geqslant 6$, there must exist three grid points that can form an isosceles triangle.