Three. (50 points) In a Cartesian coordinate system, there are 16 lattice points $(i, j)(0 \leqslant i \leqslant 3,0 \leqslant j \leqslant 3)$. If $n$ points are chosen from these 16 points, there will always be four points that form the vertices of a square. Find the minimum value of $n$.

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Three. (50 points) In a Cartesian coordinate system, there are 16 lattice points $(i, j)(0 \leqslant i \leqslant 3,0 \leqslant j \leqslant 3)$. If $n$ points are chosen from these 16 points, there will always be four points that form the vertices of a square. Find the minimum value of $n$.
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Three, the minimum value of $n$ is 11.
Below is the proof that the minimum value of $n$ is 11.
(1) There exist 10 points $(0,0)$, $(1,0)$, $(2,0)$, $(2,1)$, $(3,1)$, $(0,2)$, $(3,2)$, $(0,3)$, $(1,3)$, $(3,3)$, among which no four points form a square.
(2) In any 11 points, there must be four points that form the vertices of a square.

Now consider the situation after removing five points, leaving 11 points.
First, we prove two lemmas.
Lemma 1: In the nine grid points $(i, j) (0 \leqslant i \leqslant 2$, $0 \leqslant j \leqslant 2)$, any seven points chosen will always contain four points that form the vertices of a square.

Proof of Lemma 1: If the conclusion is not true, i.e., there exist seven points such that no four points form the vertices of a square. Since the points $(0,0)$, $(2,0)$, $(2,2)$, $(0,2)$ form the vertices of a square, at least one of these four points must be removed. Without loss of generality, assume the point $(2,2)$ is removed. The points $(1,0)$, $(1,1)$, $(2,1)$, $(2,0)$ and the points $(0,1)$, $(0,2)$, $(1,2)$, $(1,1)$ each form the vertices of a square, so the point $(1,1)$ must be removed. Among the remaining seven points, the points $(1,0)$, $(2,1)$, $(1,2)$, $(0,1)$ form the vertices of a square, which contradicts the assumption.
Lemma 1 is proved.
Lemma 2: In the 12 grid points $(i, j) (0 \leqslant i \leqslant 3$, $0 \leqslant j \leqslant 2)$, any nine points chosen will always contain four points that form the vertices of a square.

Proof of Lemma 2: Choose any nine points, i.e., remove any three points. If at least one of the three removed points is in the first column or the fourth column, then by Lemma 1, the conclusion holds. If the three removed points are in the same column (the second column or the third column), then the conclusion is obviously true. By symmetry, removing three points can be considered as removing one point from the second column and two points from the third column. To ensure that the first and second columns do not contain the vertices of a square, only the point $(1,1)$ can be removed. Thus, the remaining points $(1,0)$, $(1,2)$, $(2,0)$, $(2,2)$ form the vertices of a square.
Lemma 2 is proved.
If two points are removed from a side column or a side row, by Lemma 2, the conclusion holds.

Assume that at most one point is removed from each side column and side row. Since the points $(0,0)$, $(3,0)$, $(3,3)$, $(0,3)$ form the vertices of a square, at least one of these four points must be removed. By symmetry, assume the point $(3,3)$ is removed. The points $(1,0)$, $(3,1)$, $(2,3)$, $(0,2)$ and the points $(2,0)$, $(3,2)$, $(1,3)$, $(0,1)$ each form the vertices of a square, so one point must be removed from each, and it must not be in the same row or column as the point $(3,3)$. Therefore, the points $(1,0)$ and $(0,1)$ or $(2,0)$ and $(0,2)$ must be removed.

If the points $(1,0)$ and $(0,1)$ are removed, the points $(0,2)$, $(0,3)$, $(3,1)$, $(2,1)$ and the points $(2,0)$, $(3,0)$, $(3,1)$, $(2,1)$ each form the vertices of a square, so one point must be removed from each, and it must be internal. Thus, the points $(1,2)$ and $(2,1)$ are removed. At this point, among the remaining 11 points, the points $(1,1)$, $(2,0)$, $(3,1)$, $(2,2)$ form the vertices of a square, so the conclusion holds.
Similarly, the case where the points $(2,0)$ and $(0,2)$ are removed can be proved.
Thus, in 16 points, any 11 points chosen will always contain four points that form the vertices of a square.