6 In an $m \times n (m>1, n>1)$ chessboard $C$, each cell is filled with a number such that for any positive integers $p, q$ and any $p \times q$ rectangle, the sum of the numbers in the cells at the opposite corners is equal. If after filling in numbers in an appropriate $r$ cells, the numbers in the remaining cells are uniquely determined, find the minimum value of $r$. (5th All-Russian Mathematical
6. After filling the first row and the first column, the number table is uniquely determined, at this point, the table has only been filled with $m+n-1$ numbers. That is, when $r=m+n-1$, there exists a corresponding filling method. Below, we prove that for all valid filling methods, $r \geqslant m+n-1$. We use proof by contradiction. That is, when $r \leqslant m+n-2$, no matter how the table is filled, the number table is not uniquely determined. We use induction on $m+n$. When $m+n=4$, $m=n=2$, in a $2 \times 2$ number table, filling 2 numbers does not uniquely determine the table, so the conclusion holds. Assume the conclusion holds for $m+n=k$. Consider the case when $m+n=k+1>4$. The number table is filled with $r \leqslant m+n-2=k-1$ numbers, we need to prove that this number table is not uniquely determined. To use the assumption, we should remove one row or one column. And the row or column removed should have the following properties: (1) After removing this row, the number table should still have at least two rows (otherwise, we cannot use the induction hypothesis). For this, assume $m \leqslant n$, then $n>2$. Thus, we can remove one column. (2) After removing this column, the number of numbers in the $(m-1) \times n$ number table is no more than $(m-1)+n-1=k-2$ numbers, i.e., the removed column must contain at least one number. (3) Adding back the removed column, the number table is still not uniquely determined, which only requires that this column contains no more than one number (a sufficient condition). In fact, if a number $a$ is not uniquely determined in the sub-table, but is uniquely determined by two numbers $b$ and $c$ in the added column and a number $d$ in the same column as $a$, but this column contains only one number, then assume $c$ is newly added, then $c$ can only be determined by two numbers $e$ and $f$ in another column. Thus, $a$ can be uniquely determined by $e$, $f$, and $d$ in the sub-table. This is a contradiction. Now, we need to find a column that meets the above two conditions, i.e., this column contains exactly one number. Note that $m \leqslant n$, so the number of numbers in the chessboard $r \leqslant k-1=m+n-2 \leqslant 2n-2<2n$. Therefore, at least one column contains no more than one number. If this column contains no numbers, then this column is not uniquely determined, and the conclusion holds. If this column contains exactly one number $a$, then removing this column, the remaining numbers must be uniquely determined, otherwise, assume $b$ is not uniquely determined, then $b$ must be determined by $a$, thus $a$ and $b$ are two vertices of the same rectangle, and the other 3 numbers in this rectangle are known, among these 3 numbers, one must be in the same column as $a$, which is a contradiction. Therefore, the minimum value corresponding to the $(m-1) \times n$ number table $\leqslant r-1 \leqslant k-2=m+n-3<m+n-2$, which contradicts the induction hypothesis.