A checkboard and sums problem, prove that there is a column or row such that the absolute value of sum doesn't exceed $2018^2/2$ Now we have a $2018\times 2018$ checkboard. Each space is filled with one integer with absolute value not bigger than $2018$. 
Suppose the sum of all the numbers is $0$, prove that there is a column or a row such that the sum of $2018$ integers in that column or row is with absolute value not bigger than $2018^2/2$. 
My attempt: I let the sum of each row be $R_i$ with $i=1,2,\cdots,2018$, and let $C_j$ the same for column sum. Then I have $$\begin{cases}R_1+\cdots+R_{2018}=0\\C_1+\cdots+C_{2018}=0\end{cases}$$
Since each number does not have absolute value bigger than $2018$, I conclude that $$|R_i|\leq 2018^2,\qquad |C_j|\leq 2018^2.$$
But the problem want me to collapse to half of the bound. Any suggestion?
 A: $2018$ is just the year this problem is from (what contest?); it doesn't mean anything special here. Let's make that a $2n\times 2n$ matrix, with a bound of $m$ on the entries. The worst case scenario, in which each row and each column has the absolute value of its sum equal to $mn$: the first $n$ rows have $n$ entries equal to $m$ followed by $n$ entries equal to zero, and the last $n$ rows have $n$ entries equal to zero followed by $n$ entries equal to $-m$. A $4\times 4$ example:
$$\begin{pmatrix}m&m&0&0\\ m&m&0&0\\ 0&0&-m&-m\\ 0&0&-m&-m\end{pmatrix}$$
Why look at that? Because that's our equality case - any argument we make will have to keep that in mind.
So then, here's a hint: look at the rows with positive sum, the rows with negative sum, the columns with positive sum, and the columns with negative sum. In that equality case, the intersection of the rows with positive sum and the columns with negative sum are all zeros - can we make something from that?
All right, time for the full solution.
Suppose there are $n+r$ rows with nonnegative sum, and their elements sum to $R$. The $n-r$ rows with negative sum total to $-R$, since the sum of all elements is zero. Also, let the number of columns with nonnegative sum be $n+c$, with total sum $C$. This, of course, gives $n-c$ columns with negative sum and total sum $-C$.
Consider the sum of all elements in rows with nonnegative sum, minus the sum of all elements in columns with negative sum. The $(n+r)(n+c)$ elements with both row and column nonnegative appear in this sum with a positive sign, the $(n-r)(n-c)$ elements with both row and column negative appear with negative sign, and the elements in nonnegative rows and positive columns cancel out, not appearing in the sum.
Since each element we're adding in is between  $-m$ and $m$, we get
$$R+C \le m(n+r)(n+c)+m(n-r)(n-c)=m(2n^2+2rc)$$
On the other hand, if $A$ is the smallest absolute value of a row or column sum, we get
$$R+C \ge A(n+r)+A(n-c)=A(2n+r-c)$$
Of course, we could have done this all slightly differently. We don't have to use the same rows and columns to get our lower bound for $R+C$; it's also true that
$$R+C\ge A(2n+c-r),\quad R+C\ge A(2n+r+c),\quad R+C\ge A(2n-r-c)$$
from (pos. cols - neg. rows), (pos. rows + pos. cols), and (-neg. rows - neg. cols) respectively. Consolidating these inequalities and combining with our upper bound,
$$A(2n+|r|+|c|) \le R+C\le m(2n^2+2rc)$$
$$A \le m\cdot \frac{2n^2+2rc}{2n+|r|+|c|} \le m\cdot\frac{2n^2+n|r|+n|c|}{2n+|r|+|c|}=mn$$
We used the fact that $|r|, |c|\le n$ there to estimate the numerator; $rc\le n|r|,n|c|$.
And there it is - the smallest absolute value of a row or column sum is at most $mn$. In the originally stated version of the problem, with $m=2018$ and $n=1009=\frac{2018}{2}$, that says that there's some row or column summing to at most $\frac{2018^2}{2}$. Done.
