11. (15 points) Divide the sides of an equilateral $\triangle ABC$ with side length 3 into three equal parts, and draw lines parallel to the other two sides through each division point. The 10 points where the sides of $\triangle ABC$ and these parallel lines intersect are called grid points. If $n$ grid points are chosen from these 10 grid points, there will always be three grid points that can form an isosceles triangle (including equilateral triangles). Find the minimum value of $n$.
11. $n_{\min }=5$.

Let the two equal division points on side $AB$ from point $A$ to $B$ be $D$ and $E$, the two equal division points on side $BC$ from point $B$ to $C$ be $F$ and $G$, and the two equal division points on side $CA$ from point $C$ to $A$ be $H$ and $I$, with the central grid point being $K$.

If the minimum value of $n$ is 4, taking the grid points $A, D, E, B$, then there do not exist three grid points that can form an isosceles triangle.
Therefore, $n \geqslant 5$.
We now prove: Any selection of five grid points will definitely contain three grid points that can form an isosceles triangle.
Assume the selected points are red points.
We only need to prove: There must exist an isosceles triangle formed by red points.

If these five red points include grid point $K$, divide the other nine grid points into three sets:
$$
L=\{D, E, F\}, M=\{G, H, I\}, N=\{A, B, C\}.
$$

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 red point $K$.

If these five red points do not include grid point $K$, when grid point $A$ is a red point, in
$$
\begin{array}{l}
U=\{D, I\}, V=\{E, H\}, \\
W=\{F, G\}, T=\{B, C\}
\end{array}
$$

if one 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 $G, C$ are not red points. Thus, $F$ is a red point, and 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, thus $E$ 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, when grid point $B$ or $C$ is a red point, the conclusion still holds.

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