Answer: $B$ is divisible by $C$.

Consider any prime number $p$, and its power for each number in the table. Replace all the numbers in the table with a power of the chosen prime number. Let the $m \times n$ table is filled with $\alpha_{i,j}$, $i=1, \overline{m}$, $j=1, \overline{n}$. Now $\beta_i$ is the greatest number from the corresponding row, and $B$ is the smallest of $\beta_i$. Similarly, $\gamma_j$ is the smallest number of the corresponding column, and $\Gamma$ is the greatest of $\gamma_j$. Thus both $B$ and $\Gamma$ are in the table. If they belong to the same row or column, then $B \ge \Gamma$ by construction. If they are in different rows and columns, find $\Delta$, which is the number on the intersection of the column of $\Gamma$ and the row of $B$. Then by construction $\Gamma \le \Delta \le B$. Thus, the power of $p$ in $B$ is not less than its power in $C$. Since $p$ is arbitrary, $C|B$.