Three. (Total 50 points) In a $13 \times 13$ square grid, select the centers of $k$ small squares such that no four of these points form the vertices of a rectangle (with sides parallel to those of the original square). Find the maximum value of $k$ that satisfies the above requirement.
Three, let the $i$-th column have $x_{i}$ points $(i=1,2, \cdots, 13)$, then $\sum_{i=1}^{13} x_{i}=k$, the $x_{i}$ points in the $i$-th column form $\mathrm{C}_{x_{i}}^{2}$ different point pairs (if $x_{i}<2$, then $\mathrm{C}_{x_{i}}^{2}=0$).

If an additional column is added to the side of a $13 \times 13$ square, and each point pair is projected onto this column. Since any 4 different points are not the vertices of a rectangle, the projections of different point pairs on the newly drawn column are different. There are $C_{13}^{2}$ different point pairs on the newly drawn column, thus we get $\sum_{i=1}^{13} \mathrm{C}_{x_{i}}^{2} \leqslant \mathrm{C}_{13}^{2}$, i.e., $\sum_{i=1}^{13} x_{i}\left(x_{i}-1\right) \leqslant 13 \times 12$.
Hence $\sum_{i=1}^{13} x_{i}^{2} \leqslant 156+k$.
$\because \sum_{i=1}^{13} x_{i}^{2} \geqslant \frac{1}{13}\left(\sum_{i=1}^{13} x_{i}\right)^{2}=\frac{k^{2}}{13}$,
$\therefore k^{2} \leqslant 13 \times 156+13 k$, which gives $-39 \leqslant k \leqslant 52$.
When $k=52$, a graph that meets the conditions can be constructed (as shown in Figure 3).
In summary, the maximum value of $k$ is 52.