(11) (15 points) Divide each side of the equilateral $\triangle A B C$ with side length 3 into three equal parts, and draw lines parallel to the other two sides through each division point. The 10 points formed by the intersections of the sides of $\triangle A B C$ and these parallel lines are called grid points. If $n$ grid points are chosen from these 10 grid points, there will definitely be three grid points that can form an isosceles triangle (including equilateral triangles). Find the minimum value of $n$.
(11) Let the two equal division points on side $AB$ from $A$ to $B$ be $D$ and $E$; the two equal division points on side $BC$ from $B$ to $C$ be $F$ and $G$; and the two equal division points on side $CA$ from $C$ to $A$ be $H$ and $I$. The central grid point is $K$.

If the minimum value of $n$ is 4, taking the grid points $A$, $D$, $E$, and $B$, there does not exist three grid points that can form an isosceles triangle, hence $n \geqslant 5$.

We now prove that for any 5 selected grid points, there must exist three grid points that can form an isosceles triangle.

Assume the selected points are red points. We will prove that there must exist an isosceles triangle formed by red points.

If these 5 red points include the grid point $K$, divide the other 9 grid points into three sets $L=\{D, E, F\}$, $M=\{G, H, I\}$, and $N=\{A, B, C\}$. By the pigeonhole principle, there must be at least one set containing at least two red points. Regardless of which set and which two points are red, they will form an isosceles triangle with the red point $K$.

If these 5 red points do not include the grid point $K$, and if grid point $A$ is a red point, then consider the sets $U=\{D, I\}$, $V=\{E, H\}$, $W=\{F, G\}$, and $T=\{B, C\}$. If any of these sets contains two red points, the conclusion holds. Otherwise, each set contains exactly one red point. Assume $D$ is a red point, then $I$ is not a red point. If $B$ is a red point, then $G$ and $C$ are not red points, so $F$ is a red point. Regardless of whether $E$ or $H$ is a red point, they will form an isosceles triangle with $D$ and $F$. If $B$ is not a red point, then $C$ is a red point, so $E$ is not a red point, and $H$ is a red point. Regardless of whether $F$ or $G$ is a red point, they will form an isosceles triangle with $D$ and $H$ or $H$ and $C$. Similarly, if grid point $B$ or $C$ is a red point, the conclusion still holds.

If $K$, $A$, $B$, and $C$ are not red points, then among $D$, $E$, $F$, $G$, $H$, and $I$, there are 5 red points, and the conclusion is clearly true.