Given a positive integer $n$ greater than 2004, fill the numbers $1, 2, \cdots, n^2$ into an $n \times n$ chessboard (consisting of $n$ rows and $n$ columns of squares) such that each square contains exactly one number. If a number in a square is greater than the numbers in at least 2004 other squares in its row and at least 2004 other squares in its column, then this square is called a "super square." Determine the maximum number of "super squares" on the chessboard.
(Feng Yuefeng, problem contributor)
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 smallest numbers in each row are not "row superior," there are \( n-2004 \) "row superior" cells in each row. A cell that is "superior" 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 (where \( i=1,2, \cdots, n \)) and the \( i, i+1, \cdots, i+2003 \) (take the remainder when greater than \( n \) modulo \( n \)) columns with " * ". Fill the numbers \( 1,2, \cdots, 2004 n \) 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 "row superior," totaling \( n(n-2004) \) cells.

At this point, each row has 2004 cells with " * ", and each column also has 2004 cells with " * ". In fact, 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 \( n(n-2004) \).