Final Round of National Olympiad

Each row contains $k$ numbers less than the median and $k$ numbers greater than the median. Thus $k+1$ numbers in each row do not exceed the median of that row. In rows whose median does not exceed $m$, these $k+1$ numbers do not exceed $m$ either. There are $k+1$ such rows. Consequently, there are at least $(k+1{)}^2$ numbers in the table that do not exceed $m$. Only one of them is equal to $m$, whence $(k+1{)}^2-1=k^2+2k$ numbers are less than $m$. As $k$ is positive by assumption, we have $1<4k$ and $4k+1<4m$. Now $\frac{k^2+2k}{(2k+1{)}^2}=\frac{k^2+2k}{4k^2+4k+1}>\frac{k^2+2k}{4k^2+8k}=\frac{1}{4}$ and we are done.