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firstly we note that the characteristic can be made to be $\frac{n+1}{n}$; indeed if we consider the grid with $1,n+1,2n+1,\ldots,n^2-n+1$ along the first row(in that order), $2,n+2,\ldots,n^2-n+2$ shifted cyclically by one position along the second row( so that the row reads $n^2-n+2,2,n+2,\ldots,n^2-2n+2$) and so on(i hope the grid is clear-i am just lazy to write it out), then one can see that for this grid $\frac{n+1}{n}$ is the characteristic of this grid. now for any grid, we argue that the characteristic of the grid is $\le\frac{n+1}{n}$.note that the function $f(x):=\frac{x+n-1}{x}$ satisfies $f(x)<\frac{n+1}{n}$ iff $x>n^2-n$. hence if two of the last $n$ numbers are in the same row(column) then the characteristic of the grid is $<\frac{n+1}{n}$.so suppose that the last $n$ numbers namely $n^2-n+1,n^2-n+2,\ldots,n^2$ lie along a transversal(no two in the same row/column). now consider the position of $n^2-n$; it lies in the same row or column of one of $n^2-n+1,n^2-n+2,\ldots,n^2-1$, say $b$. then $ \frac{b}{n^2-n}\leq \frac{n^2-1}{n^2-n}=\frac{n+1}{n}$ and so the characteristic of the grid is $\le \frac{b}{n^2-n}\le \frac{n+1}{n}$.

The answer is $ \frac{n+1}{n}$. I shall first prove that the characteristic cannot be of higher value than $ \frac{n+1}{n}$. If we have two of the numbers from the set $ \{n^2-n+1,n^2-n+2,...,n^2\}$ in the same row or column then the characteristic is less than or equal to $ \frac{n^2}{n^2-n+1}<\frac{n+1}{n}$. So, if we want the characteristic to be of higher value we have to have all the $ n$ numbers from the set $ \{n^2-n+1,n^2-n+2,...,n^2\}$ in different rows and columns. Now, we must place the number $ n^2-n$ in the same row or column of exactly two of the numbers from the set $ \{n^2-n+1,n^2-n+2,...,n^2\}$ and so the maximum value for the characteristic is less than or equal to $ \frac{n^2-1}{n^2-n}=\frac{n+1}{n}$. Now, let's show this characteristic can be attained. Cover the $ n\times n$ square in the following way: Image not found. Using the inequality $ \frac{a+n}{b+n}<\frac{a}{b},a>b>0$ we see that for row $ i$ (counting rows $ 1,2,...,n$ from top to bottom) the minimum value of the ratio ($ >1$) between two numbers is $ \frac{n^2-i+1}{n^2-n-i+1}\geq \frac{n^2}{n^2-n}=\frac{n}{n-1}>\frac{n+1}{n}$. Doing almost the same for the columns (using $ \frac{a+n-1}{b+n-1}<\frac{a}{b},a>b>0$), as the difference between two numbers in the same column is at least $ n-1$, we have that the minimum value of the ratio ($ >1$) between two numbers in column $ i>1$ (counting $ 1,2,...,n$ from left to right) is $ \frac{n^2-i+1}{n^2-n-i+2}\geq \frac{n^2-1}{n^2-n}=\frac{n+1}{n}$ and in column $ 1$ is more than $ \frac{n+1}{n}$. We then conclude that the characteristic in this case is $ \frac{n+1}{n}$ and consequently this is its highest possible value.

We claim that the answer is $\frac{n+1}{n}$. Consider the numbers from $S = n^2-n\rightarrow n^2$. Then, by pigeonhole there is a pair of same-row and a pair of same-column fractions. The largest fractions made up of pairs from $S$ are in order \[\frac{n^2}{n^2-n} >\frac{n^2-1}{n^2-n} > \frac{n^2}{n^2-n+1},\ldots\]Since we are forced to have at least two fractions, we are guaranteed something $\leq $ \[\frac{n^2-1}{n^2-n}=\frac{n+1}{n}\]thus the maximum possible characteristic is $\frac{n+1}{n}$. This is achievable! We are able to keep all distances between any pair of boxes $\geq n-1$, and by avoiding $(n^2,n^2-n+1)$ pair the characteristic is $\frac{n^2}{n^2-n}>\frac{x+(n-1)}{x}$ for all legal $x$. Label the columns and rows $0\to n-1$ from top left to bottom right. Let $f(r,c)$ be the value at the point $(r,c)$. Label the diagonals TL to BR with $n^2\to n^2-n+1$. Label the left column with the point $(r,0)$ assigned $n(n-r)$. Then, increment by $(n-1)$ going to the right until you hit the diagonal. The top right triangle is assigned by labelling starting from the $1\to n-1$, labelling $(n-1)\cdot c$ in the $c$-th column. Then, write $f(r,c) = f(r-1,c-1)-1$ includegraphics[scale=0.3]{1998-A2.jpg} Thus, we have shown the characteristic is at most $\frac{n+1}{n}$ and that it is achievable so we are done.

We claim that the answer is $\boxed{\frac{n+1}{n}}$. First, we prove that every grid arrangement has a ratio less than or equal to $\frac{n+1}{n}$. Consider the $n$ highest elements in the grid ($n^2-n+1$ through $n^2$). If two of them are on the same row or column, then there will be a ratio no higher than $\frac{n^2}{n^2-n+1}$ present, and it is easy to show that this is less than $\frac{n+1}{n}$. If every one of these elements has it's own column and row, the element $n^2-n$ must share a row with one of these $n$ highest elements and share a column with another. Thus it must share a row or column with a number no greater than $n^2-1$, giving a ratio no greater than $\frac{n^2-1}{n^2-n}=\frac{(n+1)(n-1)}{n(n-1)}=\frac{n+1}{n}.$ Now we will prove that there exists an arrangement with a characteristic of $\frac{n+1}{n}$. Number the rows and columns from $0$ to $n-1$. If a cell has row number $r$ and column number $c$, give it the entry $(n-1)r-cn \pmod {n^2}$ (if this returns $0$, give the entry $n^2$). Every row consists of $n$ numbers in arithmetic sequence with common difference $n$, so that if one entry in that row has a certain residue $r$ modulo $n$, all residues modulo $n^2$ that are $r$ mod $n$ are covered. However, it is clear that column $0$ contains every residue mod $n$; thus this arrangement contains all numbers from $1$ to $n^2$. Claim: In this arrangement, there do not exist any entries in the grid in the same row or column with difference less than $n-1$. Proof. Any two elements in the same row must have the same residue mod $n$ and thus must differ by at least $n$. If two elements are in the same column, then taking off mod $n^2$ from the construction tells us we have an arithmetic sequence with common difference $n-1$ and thus mod $n^2$ two entries in the same column can only differ by any multiple of $n-1$ up to $(n-1)^2$ (if the entries are in the first and last row). Since $(n-1)^2\equiv 2n-1 \pmod {n^2}$, none of the differences are within $n-1$. $\blacksquare$ We are now in a position to prove that every ratio is greater than or equal to $\frac{n+1}{n}$. If we subtract one from all ratios, we get the difference divided by the smaller number. The difference is at least $n-1$ by the claim and thus the numerator cannot be smaller than $n-1$, in addition, as $n^2$ and $n^2-(n-1)$ don't share a row or column, the denominator cannot be greater than $n^2-n$, leading to a minimum possible ratio in this configuration to be $1+\frac{n-1}{n^2-n}=\frac{n+1}{n}$.

We claim that the answer is $\boxed{\frac{n+1}{n}}$. First, we prove that every grid arrangement has a ratio less than or equal to $\frac{n+1}{n}$. Consider the $n$ highest elements in the grid ($n^2-n+1$ through $n^2$). If two of them are on the same row or column, then there will be a ratio no higher than $\frac{n^2}{n^2-n+1}$ present, and it is easy to show that this is less than $\frac{n+1}{n}$. If every one of these elements has it's own column and row, the element $n^2-n$ must share a row with one of these $n$ highest elements and share a column with another. Thus it must share a row or column with a number no greater than $n^2-1$, giving a ratio no greater than $\frac{n^2-1}{n^2-n}=\frac{(n+1)(n-1)}{n(n-1)}=\frac{n+1}{n}.$ Now we will prove that there exists an arrangement with a characteristic of $\frac{n+1}{n}$. Number the rows and columns from $0$ to $n-1$. If a cell has row number $r$ and column number $c$, give it the entry $(n-1)r-cn \pmod {n^2}$ (if this returns $0$, give the entry $n^2$). Every row consists of $n$ numbers in arithmetic sequence with common difference $n$, so that if one entry in that row has a certain residue $r$ modulo $n$, all residues modulo $n^2$ that are $r$ mod $n$ are covered. However, it is clear that column $0$ contains every residue mod $n$; thus this arrangement contains all numbers from $1$ to $n^2$. Claim: In this arrangement, there do not exist any entries in the grid in the same row or column with difference less than $n-1$. Proof. Any two elements in the same row must have the same residue mod $n$ and thus must differ by at least $n$. If two elements are in the same column, then taking off mod $n^2$ from the construction tells us we have an arithmetic sequence with common difference $n-1$ and thus mod $n^2$ two entries in the same column can only differ by any multiple of $n-1$ up to $(n-1)^2$ (if the entries are in the first and last row). Since $(n-1)^2\equiv 2n-1 \pmod {n^2}$, none of the differences are within $n-1$. $\blacksquare$ We are now in a position to prove that every ratio is greater than or equal to $\frac{n+1}{n}$. If we subtract one from all ratios, we get the difference divided by the smaller number. The difference is at least $n-1$ by the claim and thus the numerator cannot be smaller than $n-1$, in addition, as $n^2$ and $n^2-(n-1)$ don't share a row or column, the denominator cannot be greater than $n^2-n$, leading to a minimum possible ratio in this configuration to be $1+\frac{n-1}{n^2-n}=\frac{n+1}{n}$.

If any of $\{n^2-n+1,n^2-n+2,\dots,n^2\}$ were in the same row or column, then there would be a ratio that is less than or equal to $\tfrac{n^2}{n^2-n+1}>\tfrac{n+1}{n}.$ If none of them were in the same row, then $n^2-n$ would be in the same column as one of them and the same row as anothe. Thus, there would be a ratio that is less than or equal to $\tfrac{n^2-1}{n^2-n}=\tfrac{n+1}{n}.$ Thus, the characteristic is at least $\tfrac{n+1}{n}.$ It just remains to show that a construction exists. Consider the following construction: put $1,n+1,2n+1,\dots,n^2-n+1$ in the first row, then $2,n+2,2n+2,\dots,n^2-n+2$ in the second, and so on where the row with $k,n+k,2n+k,\dots,n^2-n+k$ being cyclically shifted to the right by $k-1.$ We have \[\frac{a}{b}\ge \frac{n+1}{b}\iff \frac{b}{b-a}\le n.\]Looking at rows, the minimum difference is $n$ so $\tfrac{b}{b-a}\le \tfrac{n^2-n}{n}< n.$ Looking at columns, the minimum difference is $n-1$ so $\tfrac{b}{b-a}\le \tfrac{n^2-n+1}{n-1}\le n.$ However, we've made sure that $n^2$ and $n^2-n+1$ are not in the same line. Thus, the $\tfrac{b}{b-a}\le \tfrac{n^2-n}{n-1}=n$ as desired.

The answer is $\frac{n+1}{n}$. Achievability: This can be achieved with the grid below for $n=6$ which can easily be generalized. [asy][asy] unitsize(1.5cm); defaultpen(font("OT1","cmss","m","n")); defaultpen(fontsize(10pt)); string[][] x = { {"1","6","11","16","21"}, {"22","2","7","12","17"}, {"18","23","3","8","13"}, {"14","19","24","4","9"}, {"10","15","20","25","5"}, }; for (int i=0; i<6; ++i) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); } path p = (0,0)--(1,0)--(1,1)--(0,1)--cycle; for(int k = 0; k<5; ++k){ for(int l = 0; l<5; ++l){ if(x[k][l]!="0"){ label(scale(2)*(x[k][l]),(l+0.5,-k+4.5)); }; }; }; [/asy][/asy] We have that every row obviously has a minimum ratio of $\frac{n+1}{n}$. For the columns, note that the $k$th column is $k,n+k-1, \cdots , (k-1)n+1, (k+1)n, (k+2)n-1, \cdots , n^2-(n-k-1)$. After investigating some edge cases and the main case, we can get that this grid has every ratio bigger than $ \frac{n+1}{n}$. Bound: Assume not. Consider the numbers in $\{n^2-n+1, n^2-n+2, \cdots , n^2\}$. If two of these are in the same row or column, then their ratio is at most $\frac{n^2}{n^2-n+1}<\frac{n+1}{n}$. Thus we must have that they are all in different columns and rows. We have that $n^2-n$ must share a row or column with $2$ numbers, and the minimum must be less than or equal to $n^2-1$, so the ratio is at most $\frac{n^2-1}{n^2-n}=\frac{n+1}{n}$, so we are done.

We claim that the answer is $1+\frac{1}{n}$. We denote the cell in the $i$th row and $j$th column as $(i,j)$. Label the cell $(i,j)$ as $1-(n-1)(i-1)+n(j-1) \pmod {n^2}$ with 0 being $n^2$ (essentially going on diagonals while moving up by one each time). In such a configuration, two adjacent cells differ by at least $n-1$, and the $n^2$ cell is only adjacent to $n^2-n$, but the $n^2-1$ is adjacent to $n^2-n$, which gives a ratio of $1+\frac{1}{n}$. In order to show that a larger value is not obtainable, we will attempt to obtain it. Since $n^2-n+1$ through $n^2$ must be in different rows and columns, we place them diagonally along the grid. Consider $n^2-n$. The only cell that is not in the same row or column as a number from $n^2-n+1$ to $n^2-1$ inclusive is the cell labeled $n^2$, so $n^2-n$ must be in the same row or column as one of $n^2-n+1$ through $n^2-1$. However, this would give a ratio of at most $1+\frac{1}{n}$, so we are done.

The answer is $\frac{n+1}{n}$. To achieve this, let $a_{ij}$ be the cell in the $i$th row and $j$th column. Then \[a_{ij}\equiv 1+(i-1)n-(j-1)(n-1)\pmod {n^2}.\]To prove the bound, consider the set of $n+1$ numbers $\{n^2-0,\dots,n^2-n\}$. Clearly at least two pairs of numbers here are in the same row or column, and thus the characteristic is at most the second-largest large-to-small ratio here, which occurs at $\frac{n^2-1}{n^2-n}=\frac{n+1}{n}$.