For any square in $ S$ we can have at most $ 3$ verticies in $ T$ so we have a weak upper bound $ |T| \le \frac {3}{4}|S|$ Now, suppose there is some column $ C$ with all elements in $ T$ That is; $ C = \{(x,y)|y = 1,2,3,4\}$ and $ C\subset T$ Then the next three adjacent columns on either side of $ C$ can have at most $ 2$ elements in $ T$. Since we can easily find more efficient subsets than this we will assume that $ T$ does not contain an entire column. Therefore each column will have at most three elements in $ T$ It would be nice if we could choose three elements in every column to be an element in $ T$, making $ |T| = \frac {3}{4}|S|$. Obviously we can't, but we can have four columns of three next to each other as shown below (red dots = $ T$, blue dot = $ T'$) [asy][asy]unitsize(8); int colors[][] = {{1, 0, 0, 0}, {0, 0, 1, 0}, {0, 1, 0, 0}, {0, 0, 0, 1}}; for (int y = 0; y < 4; ++y) { for (int x = 0; x < 4; ++x) { if (colors[y][x] == 1) dot((x, -y), blue); else dot((x, -y), red); } }[/asy][/asy] Notice that if four columns of three are all adjacent then all column must be distinct permutations of red and blue (if any two were the same we would have a square). Therefore the above set up (and the reverse of it) is the only one that works. So we can pack lots of the above arrangement together if we put in another column to seperate them. [asy][asy]unitsize(8); int colors[][] = {{1, 0, 0, 0, 1, 1, 0, 0, 0}, {0, 0, 1, 0, 1, 0, 0, 1, 0}, {0, 1, 0, 0, 0, 0, 1, 0, 0}, {0, 0, 0, 1, 1, 0, 0, 0, 1}}; for (int y = 0; y < 4; ++y) { for (int x = 0; x < 9; ++x) { if (colors[y][x] == 1) dot((x, -y), blue); else dot((x, -y), red); } }[/asy][/asy] By looking at the above you should be able to see that we can only get one element from the seperating column in $ T$. Now this must be the most efficient arrangement otherwise we would need four elements in one column which we have already shown to be inefficient. So $ 1993 = 5\cdot 398 + 3$ and $ \max{|T|} = 398\cdot 13 + 3\cdot 3 = \boxed{5183}$