$10 \cdot 103$ In the coordinate plane, the set of points is given by
$$
S=\{(x, y) \mid x=1,2, \cdots, 1993, y=1,2,3,4\} .
$$

It is known that $T \subset S$ and no 4 points in $T$ are the 4 vertices of a square. Find the maximum value of $|T|$ .
[Solution] Select the point set $T_{0}$ as shown in the figure, the specific method is to select 10 points every 4 columns, because $1993=4 \times 498+1$, so $\left|T_{0}\right|=10 \times 498+3=4983$, and any 4 points in $T_{0}$ are not the 4 vertices of a square. It can be seen that the maximum value of $|T|$ is not less than 4983.

Below, we prove that for any $T \subset S,|T|=4984$, the point set $T$ cannot meet the requirements of the problem.

Lemma 1 For any 7-element subset $M$ of the set $S_{1}=\{(x, y) \mid x, y=1,2,3\}$, there are always 4 points that are the 4 vertices of a square.

It is easy to complete the proof of the lemma by noting that there is a column of 3 points all in $M$ in the set $S_{1}$. Here, we omit the details.

Lemma 2 For any 9-element subset $M$ of the set $S_{2}=\{(x, y) \mid x=1,2,3,4, y=1,2,3\}$, there are always 4 points that are the 4 vertices of a square.

Proof of Lemma 2 If not, then by Lemma 1, the first and fourth columns of the set $S_{2}$ each contain 3 points all in $M$. The other 3 points are distributed in the middle two columns, and there must be at least 2 points in one column. It is easy to see that, regardless of which column and how these two points are distributed, they will always form a square with the corresponding two points in the first or fourth column, leading to a contradiction.

Lemma 3 Let $S_{3}=\{(x, y) \mid x, y=1,2,3,4\}$, then any 11-element subset of $S_{3}$ contains 4 points that are the 4 vertices of a square.

Proof of Lemma 3 The 16 points of $S_{3}$ are arranged in a 4x4 grid. Suppose there is $M \subset S_{3}$, $|M|=11$ and no 4 points in $M$ are the 4 vertices of a square. Then by Lemma 2, each row and column of the first and fourth rows and columns of $S_{3}$ must contain at least 3 points of $M$, and among these 3 points, there is obviously at least 1 point that is a "corner point". Thus, at least 2 of the 4 corner points of $S_{3}$ are in $M$, and at most 3 are in $M$.
(1) Suppose $M$ contains exactly 2 corner points, these two corner points must be a pair of opposite vertices of the large square, let's say (1, 4) and (4, 1). Since the other two corner points are not in $M$, the other 8 points on the edges are all in $M$. Thus, the 4 points marked with * form a square, leading to a contradiction.
(2) Suppose $M$ contains exactly 3 corner points. Let's assume the bottom-right corner point is not in $M$. In this case, the distribution of points in $M$ in the first and fourth rows and columns has only two different scenarios as shown below. In each diagram, there are already 9 points. If another point is taken in the first column or the first row, it will lead to 4 points forming a square, leading to a contradiction. Therefore, the other two points of $M$ can only be taken from the 4 internal points. In either diagram, $M$ must have 1 point at the position marked with " $\times$ ", which again leads to 4 points forming a square, leading to a contradiction. This completes the proof of Lemma 3.

Returning to the proof of the original problem. Suppose there is $T \subset S,|T|=4984$, but no 4 points in $T$ are the 4 vertices of a square. By Lemma 3, any 4 consecutive columns in the set $S$ forming a $4 \times 4$ subset contain at most 10 points of $T$. Thus, each set of 4 consecutive columns contains exactly 10 points of $T$ and all 4 points in the first column are in $T$. This also leads to all 4 points in the fifth column being in $T$. In this way, the 3 columns of the 2nd, 3rd, and 4th columns still contain 6 points of $T$. Clearly, if any column contains 3 points of $T$, it will lead to 4 points forming a square. Therefore, each of the 2nd, 3rd, and 4th columns contains exactly 2 points of $T$.

It is easy to see that if the 2 points in the 2nd or 4th column are adjacent or at the top and bottom ends, it will lead to 4 points forming a square. Therefore, the 2 points in each of these columns can only be the 1st and 3rd points or the 2nd and 4th points, and the two columns must be different. Let's assume the 4 points are as shown in the figure above. In this way, no matter where the points of $T$ are in the 3rd column, it will lead to 4 points forming a square, leading to a contradiction.
In summary, the maximum value of $|T|$ is 4983.