Write $n^2$ real numbers into $n \times n$ square grid Let $k$ and $n$ be two positive integers such that $k<n$ and $k$ does not divide $n$. Show that one can fill a $n \times n$ square grid with $n^2$ real numbers such that sum of numbers in an arbitrary $k \times k$ square grid is negative, but sum of all $n^2$ numbers is positive.
Source: Homework
My idea: I presented $n = mk + r$ and write $-k^2$ into each square with coordinate of the form $(ik, jk)$ where $i, j$ are positive integers and 1 in another squares. Then the sum of each $k \times k$ square is $-1$ and sum of $n^2$ numbers is $n^2 - m^2(k^2+1)$. But this doesn't work for large $m$.
 A: This can be done so that all rows are the same.  Here's the idea of my solution: let the first column all have a value of $a > 0$, to be chosen later.  Let the next $k - 1$ columns all have $-1$ in them.  Then repeat this pattern: one column of $a$'s, $k-1$ column's of $-1$'s.  Then every $k \times k$ square has exactly one column of $a$'s and the rest are $-1$'s.  Thus, the sum over this square is $ ka - k(k-1)$.  If we want this value to be negative, then we need $a < k - 1$, i.e. $a = k-1 - \epsilon$ for some $\epsilon > 0$, to be decided later.  
Now let's look at the sum over the whole matrix.  If we write $n = mk + r$, then we have $m$ columns of $a$'s and $n - m$ columns of $-1$'s.  The total sum is then $$ nma - (n - m)n$.
Note that $n = mk + r \implies n - m = m(k - 1) + r$.  Thus, we have that the sum over the whole matrix is $$ nm((k - 1) - \epsilon) - ((k - 1)m - r)n = rn - nm\epsilon.$$
Thus, if we take $\epsilon = \frac{r}{2m}$, then we have that the sum is $$ rn - nm\epsilon = rn - nm \frac{r}{2m} = \frac{rn}{2} > 0$$
as desired.
EDIT:  The final matrix looks like: $$ \left(\begin{array}{cccc|cccc|c|ccc}
a& -1 & \cdots & -1 &a & -1 & \cdots & -1 &\cdots &a & \cdots & -1 \\
a& -1 & \cdots & -1 &a & -1 & \cdots & -1 &\cdots &a & \cdots & -1 \\
\vdots& \vdots & \ddots & \vdots & \vdots &  \vdots &  \ddots & \vdots &\ddots &\vdots & \ddots & \vdots \\
a& -1 & \cdots & -1 &a & -1 & \cdots & -1 &\cdots &a & \cdots & -1 \\
\end{array} \right)$$
Where $a = k-1 -\frac{r}{2m}$ in each block but the last I have one column of $a$'s and $(k-1)$ columns of $-1$'s.  In the last column, I have $r$ $-1$'s.  note, if I multiply the whole matrix by $2m$, then it has integer entries.
A: Here's an answer to show that in fact the OP's original idea works, if the numbers are chosen correctly.
Let our matrix take values $b-a$ at indexes $(ik,jk)$ and $b$ elsewhere, with $n$ factored as $n=mk+r$. Then the sum across a $k\times k$ submatrix is $k^2b-a$, and the sum across the whole $n\times n$ matrix is $n^2b-m^2a$. The constraints to satisfy are thus $n^2b>m^2a$ and $k^2b<a$. This implies $m^2k^2b<m^2a<n^2b$, so $mk<n$ implies $b>0$. Dividing through by $m^2b$, we get
$$k^2<\frac ab<\frac{n^2}{m^2},$$
and we can feel free to choose $b=2m^2$ and $a=n^2+m^2k^2$ (the midpoint) or any other solution which satisfies the inequality. The OP's choice used $b=1$ and $a=k^2+1$, which satisfies the first inequality, but the second inequality in the worst case has $r=1$ so $\frac nm=k+\frac1m$ and $\frac{n^2}{m^2}=k^2+2\frac km+\frac1{m^2}$, so the $+1$ is a bit too generous and is undercut for large $m$. (This can also be taken as a proof that a sufficiently large integer solution cannot take $1$ for the "background" value $b$.)
A: This is a slight variant on the construction in Marcus M's answer.
Having written $n=mk+r$ with positive integers $m$ and $0\lt r\lt k$, start with a matrix of the form
$$\pmatrix{
a&-1&\ldots&-1&\ldots&a&-1&\ldots&-1&b&\ldots&b\\
a&-1&\ldots&-1&\ldots&a&-1&\ldots&-1&b&\ldots&b\\
\vdots\\
a&-1&\ldots&-1&\ldots&a&-1&\ldots&-1&b&\ldots&b\\
}$$
where each row consists of an $a$ followed by $(k-1)$ $-1$'s, repeated $m$ times, with $r$ $b$'s at the end.  Note that the sum of any $k$ consecutive numbers in any row is either $a-(k-1)$ or $bs-(k-s)$ with $1\le s\le r$, while the sum of the entire row is $m(a-(k-1))+br$.  
Next, let $a=k-1$ and $b={k\over r}-1$.  The assumption $0\lt r\lt k$ implies $a$ and $b$ are both positive.  Our construction so far implies that the sum of any $k$ consecutive numbers in any row is either 
$$a-(k-1)=(k-1)-(k-1)=0$$ 
or 
$$bs-(k-s)=({k\over r}-1)s-(k-s)={s\over r}k-k\le0$$
while the sum of all the numbers in a row is
$$m(a-(k-1))+br=m((k-1)-(k-1))+({k\over r}-1)r=k-r\ge1$$
Thus the sum of the numbers in any $k\times k$ block is less than or equal to $0$, while the sum of the entire $n\times n$ matrix is greater than or equal to $n$.  
Now this isn't quite what's required, but it's easy to fix it:  Just subtract a very small number (anything less than $1/n$ will do) from every number in the matrix, including all the $-1$'s.  Doing so ensures that every $k\times k$ block now has a negative sum while the sum for the entire matrix is still positive.
