Four. (25 points) In a $7 \times 7$ grid, there are 64 intersection points (referred to as "nodes") where chess pieces can be placed, with at most 1 piece per point, totaling $k$ pieces. If no matter how the pieces are placed, there always exist 4 pieces such that the nodes they occupy form the four vertices of a rectangle (with sides parallel to the grid lines of the chessboard). Find the minimum value of $k$.
(Zhang Limin, problem contributor)
First, as shown in Figure 2, when $k=24$, it is possible that no 4 nodes with chess pieces form the four vertices of a rectangle, so $k \geqslant 25$.

Second, we will prove that when $k=25$, there must exist 4 nodes with chess pieces that form the four vertices of a rectangle.

If not, consider the row (or column) with the most chess pieces among all rows and columns. Since swapping any two rows (or two columns) does not affect the result, we can assume without loss of generality that the first row has the most chess pieces.
(1) When the first row has 8 chess pieces, because
$$
25-8=17,17=8 \times 2+1 \text {, }
$$

there must be another row with 2 chess pieces. Thus, the 4 chess pieces in these two rows form a rectangle, which is a contradiction.
(2) When the first row has 7 chess pieces, if no rectangle exists, then in the remaining 7 rows of the 7 columns with chess pieces, there can be at most 7 chess pieces, totaling at most 21 chess pieces, which is a contradiction.
(3) When the first row has 6 chess pieces, if no rectangle exists, then in the 6 columns with these 6 chess pieces, there can be at most $7+6=13$ chess pieces. In the remaining 2 columns, there are 12 chess pieces $(12=7+5)$, and there must be two rows with chess pieces in these two columns, forming a rectangle with the 4 chess pieces, which is a contradiction.
(4) When the first row has 5 chess pieces, if no rectangle exists, these 5 chess pieces in the 5 columns can have at most $5+7=12$ chess pieces. In the remaining 7 rows and 3 columns, there are 13 chess pieces $(13=3 \times 4+1)$. Thus, there must be one column with 5 chess pieces. The 5 chess pieces in this column can have at most 7 chess pieces in the other 3 columns, leaving 6 chess pieces in a $2 \times 3$ grid, which must form a rectangle with 4 points, which is a contradiction.
(5) When the first row has 4 chess pieces, if no rectangle exists, these 4 chess pieces in the 4 columns can have at most $4+7=11$ chess pieces. The remaining 14 chess pieces are placed in 7 rows and 4 columns. Since $14=4 \times 3+2$, there must be one column with 4 chess pieces. Thus, the 4 chess pieces in this column can have at most $4+3=7$ chess pieces in the other 4 columns, leaving 7 chess pieces in 3 rows and 3 columns, which still form a rectangle with 4 points, which is a contradiction.
In summary, the minimum value of $k$ is 25.