24.3.13 Xuan Tai Given a positive integer $n$ greater than 2004, fill the numbers $1,2,3, \cdots, n^{2}$ into the squares of an $n \times n$ chessboard (consisting of $n$ rows and $n$ columns) such that each square contains exactly one number. If a square contains a number that is greater than the numbers in at least 2004 squares in its row and at least 2004 squares in its column, then this square is called a “super square”. Find the maximum number of “super squares” on the chessboard.
For the convenience of narration, if a number in a cell is greater than the numbers in at least 2004 other cells in its row, then this cell is called "row superior". Since the 2004 cells with the smallest numbers in each row are not row superior, there are $n-2004$ row superior cells in each row. A cell that is a "superior cell" must be row superior, so the number of "superior cells" on the chessboard is no more than $n(n-2004)$.

On the other hand, fill the cells in the $i$-th row ($i=1,2,3, \cdots, n$) and the $i, i+1, \cdots, i+2003$ columns (with indices greater than $n$ taken modulo $n$) with "*". Fill the numbers $1,2,3, \cdots, 2004 n$ into the cells with "*", and fill the remaining numbers into the cells without "*". The numbers in the cells without "*" are greater than any number in the cells with "*", so all the cells without "*" on the chessboard are "superior cells", totaling $n(n-2004)$.

At this point, each row has 2004 cells with "*", and each column also has 2004 cells with "*". Specifically, when $1 \leqslant i \leqslant 2003$, the $i$-th column has "*" in the rows $1,2, \cdots, i, n+i-2003, n+i-2002, \cdots, n$. When $i \geqslant 2004$, the $i$-th column has "*" in the rows $i-2003, i-2002, \cdots, i$. Therefore, each row has 2004 cells with "*", and each column also has 2004 cells with "*".

Thus, the maximum number of "superior cells" on the chessboard is $n(n-2004)$.