The numbers from 1 to $n^2$  are randomly arranged in the cells of a $n \times n$ square ($n \geq 2$). For any pair of numbers situated on the same row or on the same column the ratio of the greater number to the smaller number is calculated.  Let us call the [b]characteristic[/b] of the arrangement the smallest of these $n^2\left(n-1\right)$ fractions. What is the highest possible value of the characteristic ?
1. **Claim:** The highest possible value of the characteristic is \(\boxed{\frac{n+1}{n}}\).

2. **Upper Bound Proof:**
   - Consider the \(n\) highest elements in the grid, which are \(n^2 - n + 1\) through \(n^2\).
   - If two of these elements are in the same row or column, the ratio of the greater to the smaller number is at most \(\frac{n^2}{n^2 - n + 1}\).
   - We need to show that \(\frac{n^2}{n^2 - n + 1} < \frac{n+1}{n}\):
     \[
     \frac{n^2}{n^2 - n + 1} < \frac{n+1}{n}
     \]
     \[
     n^2 < (n+1)(n^2 - n + 1)
     \]
     \[
     n^2 < n^3 - n^2 + n + n^2 - n + 1
     \]
     \[
     n^2 < n^3 + 1
     \]
     This inequality is always true for \(n \geq 2\).

   - If each of these \(n\) highest elements is in a different row and column, then the element \(n^2 - n\) must share a row or column with one of these \(n\) highest elements.
   - The ratio involving \(n^2 - n\) and one of the \(n\) highest elements is at most \(\frac{n^2 - 1}{n^2 - n}\):
     \[
     \frac{n^2 - 1}{n^2 - n} = \frac{(n+1)(n-1)}{n(n-1)} = \frac{n+1}{n}
     \]

3. **Existence of Arrangement with Characteristic \(\frac{n+1}{n}\):**
   - Number the rows and columns from \(0\) to \(n-1\).
   - Assign the entry in the cell at row \(r\) and column \(c\) the value \((n-1)r - cn \pmod{n^2}\). If this value is \(0\), assign \(n^2\).
   - Each row consists of \(n\) numbers in arithmetic sequence with common difference \(n\).
   - Each column contains every residue mod \(n\), ensuring all numbers from \(1\) to \(n^2\) are present.

4. **Claim:** In this arrangement, no two entries in the same row or column have a difference less than \(n-1\).
   - **Proof:**
     - Any two elements in the same row have the same residue mod \(n\) and differ by at least \(n\).
     - In the same column, the construction implies an arithmetic sequence with common difference \(n-1\). Thus, differences are multiples of \(n-1\) up to \((n-1)^2\).
     - Since \((n-1)^2 \equiv 2n-1 \pmod{n^2}\), no differences are within \(n-1\).

5. **Minimum Ratio Calculation:**
   - The difference between any two numbers in the same row or column is at least \(n-1\).
   - The smallest possible ratio is:
     \[
     1 + \frac{n-1}{n^2 - n} = \frac{n+1}{n}
     \]