Example 7 Let $m, n$ be positive integers, $m<2001, n<2002$. There are $2001 \times 2002$ different real numbers, which are filled into the squares of a $2001 \times 2002$ chessboard, so that each square contains exactly one number. If the number in a certain square is less than at least $m$ numbers in its column and also less than at least $n$ numbers in its row, then this square is called a "bad square". For all filling methods, find the minimum value of the number of bad squares $S$. (2002
Analysis and Solution: Consider a special case: Fill the $2001 \times 2002$ chessboard with the numbers $1, 2, 3, \cdots, 2001 \times 2002$ in natural order (as shown in the table below), at this time the number of bad cells $S = (2001 - m)(2002 - n)$.
\begin{tabular}{|c|c|c|c|}
\hline 1 & 2 & $\cdots$ & 2002 \\
\hline 2003 & 2004 & $\cdots$ & 4004 \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline $2000 \times 2002 + 1$ & $2000 \times 2002 + 2$ & $\cdots$ & $2001 \times 2002$ \\
\hline
\end{tabular}

We conjecture that the minimum value of $S$ is $(2001 - m)(2002 - n)$. Generally, for a $p \times q$ chessboard (where $m < p, n < q$), the number of bad cells is no less than $(p - m)(q - n)$.

We prove this by induction on $p + q$. When $p + q = m + n + 2$, since $p \geq m + 1$ and $q \geq n + 1$, we have $p = m + 1$ and $q = n + 1$. Because the cell containing the smallest number must be a bad cell, the number of bad cells is no less than $1 = (p - m)(q - n)$, so the conclusion holds. Assume that the conclusion holds for $p + q = t$. Consider the case when $p + q = t + 1$. By the lemma, assume there exists a row that does not contain any good cells in bad columns. Remove this row, and we get a $(p-1) \times q$ chessboard. Since $p + q - 1 = t$, by the induction hypothesis, this chessboard has no less than $(p-1 - m)(q - n)$ bad cells. Adding back the removed row, the original bad cells remain bad, and the bad cells in this row must also be bad (since this row does not contain any good cells in bad columns). There are $(q - n)$ bad cells in this row, so the total number of bad cells is no less than $(p-1 - m)(q - n) + (q - n) = (p - m)(q - n)$. By the principle of induction, the proposition is proved.

For the original problem, the number of bad cells is no less than $(2001 - m)(2002 - n)$. Combining this with the previous construction, we know that the minimum number of bad cells is $(2001 - m)(2002 - n)$.