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,ans. n/2,construction is easy,and to prove ans i use ideas of sub. desciption

Let $ n = 2k$, and let $ a_{ij}$ be the entry in the $ i$th row and $ j$th column. Then we claim that $ C = k$. We know that $ C \ge k$, since we can set $ a_{ij}$ to be 1 if $ i, j \le k$, -1 if $ i, j > k$, and 0 otherwise; this construction has every row and column sum having absolute value $ k$. We will now show that $ C = k$ works. Suppose for sake of contradiction that we have a matrix whose row and column sums all have absolute value more than $ k$. WLOG we can assume that the first $ a$ row sums and first $ b$ column sums are positive, while the last $ 2k - a$ and $ 2k - b$ row and column sums are non-positive (this can be achieved simply by a permutation of the rows and columns). Furthermore, we can assume that $ a + b \ge 2k$, or else we may exchange $ a$ and $ b$ with $ 2k - a$ and $ 2k - b$ and flip all signs. Define $ P = \sum_{i \le a, j \le b} a_{ij}$; $ N = \sum_{i > a, j > b} a_{ij}$; $ X = \sum_{i \le a, j > b} a_{ij}$; $ Y = \sum_{i > a, j \le b} a_{ij}$. Since $ P + X$ is the sum of the first $ a$ row sums, it is strictly greater than $ ka$. Since $ P + Y$ is the sum of the first $ b$ column sums, it is strictly greater than $ kb$. Therefore, $ P + X + P + Y > k(a + b)$. Noting that $ P + X + Y + N = 0$, we have $ P - N > k(a + b)$. There are only $ ab$ entries in the sum by which we calculate $ P$, and there are only $ (2k - a)(2k - b)$ entries for the sum for $ N$. Thus, $ k(a + b) < P - N \le ab + (2k - a)(2k - b)$ $ k(a + b) < 4k^2 - 2k(a + b) + 2ab$ $ 0 < 4k^2 - 3k(a + b) + 2ab$ $ k^2 < 9k^2 - 6k(a + b) + 4ab = (3k - 2a)(3k - 2b)$ $ 4k^2 < (6k - 2a - 2b)^2 \le (2k)^2 = 4k^2$, a contradiction. (Recall that $ 4k \ge a + b \ge 2k$).

Let $n=2k$, then $C=k$. Construction is easy, so let's prove it. In this solution, include 0 in positive numbers. When $k=1$, there exists a row or a column where the two entries have opposite sign. If not, all entries must have the same sign and they must all be 0, which is a trivial case. Sum of two opposite numbers where each has absolute value not bigger than 1 yields a number also having absolute value not bigger than 1. Here, notice that if we have one column not bigger than 1 in absolute value, the other column also satisfies this. If $k>1$ let the matrix be $M$. first, since switching rows or columns does not affect the problem, switch the order of rows and the order of columns so that both row sums and column sums are arranged biggest to smallest. (row sum : sum of entries in the row, column sum : sum of entries in the column) Now let $A= \sum_{1 \leq i,j \leq k} a_{ij}, B= \sum_{1 \leq i \leq k < j} a_{ij}, C= \sum_{1 \leq j \leq k<i} a_{ij}, D= \sum_{i, j>k} a_{ij}$, and make a 2-by-2 matrix $M'$ with entries $A, B, C, D$, i.e., merge the four k-by-k squares in $M$. Note that $|A|, |B|, |C|, |D| \leq k^2$ and $A+B+C+D=0$, so considering the $\frac{1}{k^2} M'$, by the result of case when $k=1$, we can assume $|A+C| \leq k^2$ and $|B+D| \leq k^2$, i.e. the sum of the column sums of first half columns and second half columns of $M'$ does not exceed $k^2$. Since the column sums were arranged biggest to smallest, either all column sums are positive in the first half or all column sums are negative in the second half. Let it be the first half since we can consider $-M$ if it was not the case. k column sums have sum smaller than $k^2$, so one of them must be smaller than $k$, and since it's positive, it's absolute value is smaller than k.

orl wrote: Consider a matrix of size $n\times n$ whose entries are real numbers of absolute value not exceeding $1$. The sum of all entries of the matrix is $0$. Let $n$ be an even positive integer. Determine the least number $C$ such that every such matrix necessarily has a row or a column with the sum of its entries not exceeding $C$ in absolute value. Divide the matrix into four $n/2 \times n/2$ quarters. Fill the top left with $1$s, the top right and bottom left with $0$s and the bottom right with $-1$s. Thus we get $C \ge n/2.$ We now show that $C=n/2.$ Assume on the contrary that there exists a matrix $A=(a_{ij})_{i,j=1}^n$ such that every row and column sum is either greater than $n/2$ or less than $-n/2.$ Let $R_i$ represent the $i$ th row sum and $C_i$ the $i$ th column sum. Permute the rows so that all the "positive" rows are on the top of the "negative" rows. Represent the positive ones by $R^+$ and negative by $R^-.$ Similarly permute the columns. Also, let $|R^+|$ denote the number of positive rows while $|R^-|$ the negative. Note that the sum $=0$ condition implies that the sum of the positive rows equals the magnitude of the sum of the negative rows. Claim: $|R^+|=|R^-|=n/2.$ Similarly $|C^+|=|C^-|=n/2$. Proof: If $|R^+|>n/2,$ then $|R^-|<n/2.$ Note that $R^+>n/2$ and $-n \le R^- <-n/2$ by our assumption. Hence $$\left |\sum R^- \right | \le \frac{n}{2}|R^-|<\frac{n^2}{4} < \frac{n}{2} |R^+|<\sum R^+$$a contradiction. The other case is similar. $\square$ Note that sum of all the elements is given by the sum of the top $n/2$ rows and left $n/2$ colums, minus the top left $n/2 \times n/2$ square (which is double counted before) plus the bottom right $n/2 \times n/2$ square. These top rows and left colums have a positive sum by the claim, while the latter two have the maximum, minimum value when all the member are $1, -1$ respectively. Hence Thus, \begin{align*} \sum_{i,j} a_{ij} =\sum_{i<n/2} R_i+\sum_{i<n/2} C_i +\sum_{i,j>n/2} a_{ij} -\sum_{i,j<n/2} a_{ij} >\left( \frac{n}{2} \right) \left( \frac{n}{2} \right)+\left( \frac{n}{2} \right) \left( \frac{n}{2} \right)-\frac{n^2}{4}-\frac{n^2}{4}=0 \end{align*}a contradiction. $\blacksquare$ Edit: As pointed out below, this solution has a fault; the lemma is wrong. I believe that it can be fixed, but I am yet to try it.

Wizard_32 wrote: orl wrote: Consider a matrix of size $n\times n$ whose entries are real numbers of absolute value not exceeding $1$. The sum of all entries of the matrix is $0$. Let $n$ be an even positive integer. Determine the least number $C$ such that every such matrix necessarily has a row or a column with the sum of its entries not exceeding $C$ in absolute value. Divide the matrix into four $n/2 \times n/2$ quarters. Fill the top left with $1$s, the top right and bottom left with $0$s and the bottom right with $-1$s. Thus we get $C \ge n/2.$ We now show that $C=n/2.$ Assume on the contrary that there exists a matrix $A=(a_{ij})_{i,j=1}^n$ such that every row and column sum is either greater than $n/2$ or less than $-n/2.$ Let $R_i$ represent the $i$ th row sum and $C_i$ the $i$ th column sum. Permute the rows so that all the "positive" rows are on the top of the "negative" rows. Represent the positive ones by $R^+$ and negative by $R^-.$ Similarly permute the columns. Also, let $|R^+|$ denote the number of positive rows while $|R^-|$ the negative. Note that the sum $=0$ condition implies that the sum of the positive rows equals the magnitude of the sum of the negative rows. Claim: $|R^+|=|R^-|=n/2.$ Similarly $|C^+|=|C^-|=n/2$. Proof: If $|R^+|>n/2,$ then $|R^-|<n/2.$ Note that $R^+>n/2$ and $-n \le R^- <-n/2$ by our assumption. Hence $$\left |\sum R^- \right | \le \frac{n}{2}|R^-|<\frac{n^2}{4} < \frac{n}{2} |R^+|<\sum R^+$$a contradiction. The other case is similar. $\square$ Note that sum of all the elements is given by the sum of the top $n/2$ rows and left $n/2$ colums, minus the top left $n/2 \times n/2$ square (which is double counted before) plus the bottom right $n/2 \times n/2$ square. These top rows and left colums have a positive sum by the claim, while the latter two have the maximum, minimum value when all the member are $1, -1$ respectively. Hence Thus, \begin{align*} \sum_{i,j} a_{ij} =\sum_{i<n/2} R_i+\sum_{i<n/2} C_i +\sum_{i,j>n/2} a_{ij} -\sum_{i,j<n/2} a_{ij} >\left( \frac{n}{2} \right) \left( \frac{n}{2} \right)+\left( \frac{n}{2} \right) \left( \frac{n}{2} \right)-\frac{n^2}{4}-\frac{n^2}{4}=0 \end{align*}a contradiction. $\blacksquare$ i feel there is a flaw in ur solution.why is |R+| and |R-| same. i feel that this is not always true

The number of rows with positive sum is equal to the number of rows with negative sum. I guess you got confused about the notation. If you find a flaw or a counterexample, then kindly let me know.

yes this is not always true .the number of rows with positive sum is not always equal to number of rows with negative sum.here is counterexample for n=10. 0 0 0 0 1 1 1 1 1 1 ---(sum 6>10/2=5) 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 0 0 0 1 1 1 1 1 1 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 0 -1 -1 -1 -1 -1 -1 -1 -1 -1 so this counterexample is valid.

Wizard_32 wrote: The number of rows with positive sum is equal to the number of rows with negative sum. I guess you got confused about the notation. If you find a flaw or a counterexample, then kindly let me know. i feel u messed up with the inequality of sigma|R-| to prove that the sum of matrix is not zero actually the max value of mod of the sum negative row is n but u have took it as n/2 so the inequality condition goes wrong

We claim that $C=\tfrac{N}{2},$ achieved by splitting the matrix into the four equal square submatrices $\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}.$ Note that we can permute the rows and columns such that only the first $w$ columns and the first $y$ columns have positive sum. Let $x=N-w,z=N-y.$ Let the corresponding sums of the entries in each region be $W,X,Y,Z$ as shown: [asy][asy] size(6cm); draw((0,0)--(5,0)--(5,5)--(0,5)--(0,0)); draw((2.7,0)--(2.7,5)); draw((0,2.1)--(5,2.1)); label("$w$",(0,5)--(2.7,5),N); label("$x$",(2.7,5)--(5,5),N); label("$y$",(0,5)--(0,2.1),W); label("$z$",(0,2.1)--(0,0),W); label("$W$",(1.35,3.55)); label("$X$",(3.85,3.55)); label("$Y$",(1.35,1.05)); label("$Z$",(3.85,1.05)); [/asy][/asy] So we easily get \[W+Y\geq wC,\ X+Z\leq -xC,\ W+X\geq yC,\ Y+Z\leq -zC.\]But for instance $X+Z\leq -xC\implies W+Y\geq xC,$ and similar inequalities give \[W+Y\geq \max(w,x)C,\ W+X\geq \max(y,z)C,\ Y+Z\leq -NC+\min(y,z)C,\ X+Z\leq -NC+\min(w,x)C.\]Summing in pairs, we get \[2W+X+Y\geq C(\max(w,x)+\max(y,z))\geq C(\max(w,x)-\min(w,x)+\max(y,z)-\min(y,z))+2NC+X+Y+2Z\]and the extreme inequalities gives \[2(W-Z)\geq C(|w-x|+|y-z|+2N).\]But $W-Z\leq wy+xz=wy+(N-w)(N-y)$ and $|w-x|+|y-z|=|2w-N|+|2y-N|,$ so we get \[wy+(N-w)(N-y)\geq C(|2w-N|+|2y-N|+2N).\]This rearranges to \[(2w-N)(2y-N)+N^2\geq C(|2w-N|+|2y-N|+2N).\]Finally, noting that $(2w-N)(2y-N)\leq |(2w-N)(2y-N)|$ and letting $a=|2w-N|,b=|2y-N|,$ we get \[C\leq \frac{ab+N^2}{a+b+2N}\qquad (\spadesuit).\]We show that $\tfrac{ab+N^2}{a+b+2N}\leq \tfrac{N}{2}.$ This is equivalent to $(2a-N)(2b-N)\leq N^2,$ which is true since \[-N\leq 2w-N,2y-N\leq N\implies 0\leq a,b\leq N\implies -N\leq 2a-N,2b-N\leq N.\]Thus we conclude $C\leq \tfrac{N}{2}$ from $(\spadesuit).$

let $n=2k$ we claim thet the answer $k$ the construction is easy we will prove that $C=k$ work WLOG let the first $x$ row are positive the next $y$ are negative and do the same for columns as the picture [asy][asy] size(6cm); draw((0,0)--(5,0)--(5,5)--(0,5)--(0,0)); draw((2.7,0)--(2.7,5)); draw((0,2.1)--(5,2.1)); label("$w$",(0,5)--(2.7,5),N); label("$z$",(2.7,5)--(5,5),N); label("$x$",(0,5)--(0,2.1),W); label("$y$",(0,2.1)--(0,0),W); [/asy][/asy] lemma: if we have a matrix $2 \times 2$ and $|a_{ij}| \le m$ and the sum of all entries is zero thene there exists a row or column with the sum of its entries not exceeding $m$ in absolute value. it's obvious $\blacksquare$ now split the matrix into four identical matrices [asy][asy] size(6cm); draw((0,0)--(5,0)--(5,5)--(0,5)--(0,0)); draw((2.5,0)--(2.5,5)); draw((0,2.5)--(5,2.5)); label("$W$",(1.35,3.55)); label("$X$",(3.85,3.55)); label("$Y$",(1.35,1.05)); label("$Z$",(3.85,1.05)); [/asy][/asy] from the lemma suppose $|W+X| \le k^2$ but $W+X=-Y-Z \implies |Y+Z| \le k^2$ if $x>k$ then we're done from the PHP becuase $r_1+r_2....r_{k} \le k^2$ if $x < k$ then $ y >k$ just do the same and we win

What if $n$ is odd?

The answer is $\tfrac{n}2$ with every in the top left quadrant $1$, all bottom right quadrant $-1$ and the rest are zero. Assume on the contrary that we can have every row/column sum greater than $\tfrac{n}2$ or less than $-\tfrac{n}{2}$. We can rearrange rows and columns so arrange them in descending order from left to right, from top to bottom. Let $A$ be sum of top left quadrant, $B$ be top right, $C$ be bottom left, $D$ be bottom right. WLOG suppose there are as many positive-summed rows as negative-summed ones, so the region of $A+B$ has all positive rows sums, so $A+B\ge \tfrac{n^2}{4}$. Since each column in the region $A+B$ sums to at most $\tfrac{n}2$, there are at least $\tfrac{n}2$ nonnegative sums. If the sum is nonnegative no amount of negative numbers can make the entire column a negative enough sum. Thus, $A+C\ge \tfrac{n^2}{4}$. We have \[n^2=A+B+C+D=(A+B)+(A+C)-(A-D)> \frac{n^2}{4}+\frac{n^2}{4}-(A-D)\]so $A-D> \tfrac{n^2}{2}$, absurd.

Ali3085 wrote: let $n=2k$ we claim thet the answer $k$ the construction is easy we will prove that $C=k$ work WLOG let the first $x$ row are positive the next $y$ are negative and do the same for columns as the picture [asy][asy] size(6cm); draw((0,0)--(5,0)--(5,5)--(0,5)--(0,0)); draw((2.7,0)--(2.7,5)); draw((0,2.1)--(5,2.1)); label("$w$",(0,5)--(2.7,5),N); label("$z$",(2.7,5)--(5,5),N); label("$x$",(0,5)--(0,2.1),W); label("$y$",(0,2.1)--(0,0),W); [/asy][/asy] lemma: if we have a matrix $2 \times 2$ and $|a_{ij}| \le m$ and the sum of all entries is zero thene there exists a row or column with the sum of its entries not exceeding $m$ in absolute value. it's obvious $\blacksquare$ now split the matrix into four identical matrices [asy][asy] size(6cm); draw((0,0)--(5,0)--(5,5)--(0,5)--(0,0)); draw((2.5,0)--(2.5,5)); draw((0,2.5)--(5,2.5)); label("$W$",(1.35,3.55)); label("$X$",(3.85,3.55)); label("$Y$",(1.35,1.05)); label("$Z$",(3.85,1.05)); [/asy][/asy] from the lemma suppose $|W+X| \le k^2$ but $W+X=-Y-Z \implies |Y+Z| \le k^2$ if $x>k$ then we're done from the PHP becuase $r_1+r_2....r_{k} \le k^2$ if $x < k$ then $ y >k$ just do the same and we win I believe that it is not true that all quadrants have sum less than k^2.

your brain on smoothing The answer is $n/2$, achieved by splitting the matrix into four $n/2 \times n/2$ submatrices and filling the top left with $1$ and the bottom right with $-1$. We now show that $C>n/2$ is impossible. Suppose that there was some $n \times n$ matrix with every row/column having absolute value greater than $n/2$. Since every row/column also has absolute value at most $n$, it follows that there are at least $n/3$ rows with positive sum, and at least $n/3$ rows with negative sum (same for columns). Now WLOG permute the matrix so that the rows are sorted in decreasing order from top to bottom, and the columns are sorted in decreasing order from left to right. Consider the top-left region, formed by the intersection of the positive rows and positive columns. If some entry in this region is not $1$, but we can find some other entry in its row lying in the top-right region, formed by the intersection of the positive rows and the negative columns, which is not $-1$, then we can increase the first entry and decrease the second at the same rate until some entry's absolute value reaches $1$: this clearly does not affect the minimum absolute value of any row/column nor change the signs of any row/column, and the sum of matrix entries should be the same, so we may apply this smoothing operation freely until we cannot (clearly this will terminate). When this process terminates, I claim that the entire top-left region is filled with $1$s. Indeed, if there is still some term in the top-left region which is less than $1$, then all the terms in its row which are in the top-right region must be $-1$. Since there are at least $n/3$ columns in this top-right region, it follows that the sum of the term's row is at most $-n/3+2n/3<n/2$, and since it must be positive by definition we obtain a contradiction. We may repeat the same process with the bottom-right and bottom-left region, so now suppose the top-left and bottom-right regions are filled with $1$'s and $-1$'s respectively. Let $a,b$ be the number of rows and columns respectively in the top-left region, and WLOG suppose that $a+b \geq n$ (otherwise apply the same argument to the bottom-right). Suppose the sum of values in the top-right region and bottom-left region are $x,y$ respectively. By expectation, we should have $$b+\frac{x}{a}>\frac{n}{2} \implies x>\frac{na}{2}-ab \text{ and } a+\frac{y}{b}>\frac{n}{2} \implies y>\frac{nb}{2}-ab.$$On the other hand, since the sum of the terms in the matrix is $0$, we should have $$x+y=(n-a)(n-b)-ab=n^2-na-nb \implies n^2-na-nb>\frac{na}{2}+\frac{nb}{2}-4ab\geq \frac{n(a+b)}{2}-(a+b)^2 \implies (2n-(a+b))(n-(a+b))>0,$$where we use $(a+b)^2 \geq 4ab$ (AM-GM), but this is clearly impossible for $2n\geq a+b \geq n$: contradiction. $\blacksquare$

Let $n=2k$(why is the problem not phrased this way lol). Then the answer is $\boxed{k}$, achieved by the top left quarter grid filled with $1$s and bottom right quarter grid filled with $-1$s. Now suppose that we could achieve $C>k$. Then at least $k$ of the rows must have the same sum(by pigeonhole). WLOG assume these are positive and the first $k$ rows. Since the sum of these rows is greater than $k^2$, at least $k$ of these columns have their first $k$ rows sum up to a positive number(since each is at most $k$). But since each column has absolute value at least $C$, this means that those $k$ columns have a positive sum(they don't have eonugh cells to get to $-k$). Assume WLOG that these columns are the first $k$ columns. Now consider the sum of the entire grid. The top half and the left half of the board are both greater than $k^2$, but by PIE their intersection and the outside can change the sum by at most $k^2$, meaning that the sum of the board is positive, contradiction. $\blacksquare$