Let $S = \{(x,y) \mid x = 1, 2, \ldots, 1993, y = 1, 2, 3, 4\}$. We aim to find the maximum possible value of $|T|$ for a subset $T \subset S$ such that there are no squares in $T$.

To solve this, we need to ensure that no four points in $T$ form the vertices of a square. The key observation is that for any square in $S$, we can have at most 3 of its vertices in $T$. This gives a weak upper bound:
$$|T| \leq \frac{3}{4} |S|.$$

We will use a more refined approach to maximize $|T|$. Consider the columns of $S$. If a column $C$ contains all its elements in $T$, then the adjacent columns can have at most 2 elements in $T$ to avoid forming squares. Thus, it is more efficient to avoid having all elements of any column in $T$.

We can choose 3 elements from each column in $T$ without forming squares. To achieve this, we can use a pattern where each set of 4 adjacent columns has distinct permutations of 3 elements in $T$ and 1 element not in $T$. This pattern avoids forming squares and maximizes the number of elements in $T$.

For example, consider the following arrangement for 4 columns:
$$\begin{array}{cccc} \bullet & \circ & \circ & \circ \\ \circ & \circ & \bullet & \circ \\ \circ & \bullet & \circ & \circ \\ \circ & \circ & \circ & \bullet \\ \end{array}$$
Here, $\bullet$ represents an element in $T$ and $\circ$ represents an element not in $T$.

This pattern can be repeated, with a separating column containing only 1 element in $T$ to avoid forming squares. Given that there are 1993 columns, we can divide them into groups of 5 columns (4 columns with 3 elements each and 1 separating column with 1 element).

Thus, we have:
$$1993 = 5 \cdot 398 + 3.$$

The maximum number of elements in $T$ is:
$$398 \cdot 13 + 3 \cdot 3 = 5183.$$

Therefore, the maximum possible value of $|T|$ is:
$$\boxed{5183}.$$