6. (2002 Vietnam Mathematical Olympiad) Given positive integers $m, n, m<2001, n<2002$, there are $2001 \times 2002$ different real numbers. Place these numbers into a $2001 \times 2002$ rectangular chessboard (the chessboard has 2001 rows and 2002 columns), so that each cell contains one number, and each number is placed in one cell. A cell is called a "bad cell" if the number in it is less than at least $m$ numbers in its column and less than at least $n$ numbers in its row. For each placement, let $S$ be the number of "bad cells". Find the minimum value of $S$.
6. Replace $2001, 2002$ with $p, q$ ($m \leqslant p, n \leqslant q$), and strengthen the problem to prove by induction on $p+q$ that
$S \geqslant (p-m)(q-n)$.
Clearly, when $p+q=2,3,4$, inequality (1) holds.
Assume that when $p+q=k$, inequality (1) holds.
When $p+q=k+1$, consider a $(p, q)$ chessboard. If $m=p$ or $n=q$, inequality (1) clearly holds. Now consider the case where $m<p$ and $n<q$. If a number in a cell is less than at least $n$ numbers (or $m$ numbers) in its row (or column), then the cell is called a bad row cell (or bad column cell).
In a $(p, q)$ chessboard, if every bad column cell is also a bad row cell, and every bad row cell is also a bad column cell, then it is easy to see that
$S=(p-m)(q-n)$.
Therefore, inequality (1) holds.
Assume that in a $(p, q)$ chessboard, there exists a cell that is only a bad row cell or a bad column cell. Let $a$ be the smallest number written in such a cell, and assume without loss of generality that $a$ is a bad row cell. Then, all $(p-m)$ bad column cells in the column containing $a$ are smaller than $a$, and thus must be bad cells in the chessboard. Removing the column containing $a$, the $(p, q-1)$ chessboard has cells that are bad cells in the previously assumed $(p, q)$ chessboard. Since $p+q-1=k$, by the induction hypothesis, the number of bad cells in the $(p, q-1)$ chessboard is at least $(p-m)(q-1-n)$. Therefore, the number of bad cells in the $(p, q)$ chessboard is at least
$(p-m)(q-1-n)+(p-m)=(p-m)(q-n)$.
Thus, inequality (1) holds.
Now, in a $(p, q)$ chessboard, give a placement such that the number of bad cells equals $(p-m)(q-n)$: place the $p \times q$ numbers in increasing order from left to right in the cells.
Thus, $S_{\text{min}}=(p-m)(q-n)$, and the required minimum value is
$(2001-m)(2002-n)$.