Three. (50 points) Arrange the $n^2$ natural numbers $1, 2, \cdots, n^2$ (where $n \geqslant 2$) randomly in an $n \times n$ square grid. For any two numbers in the same row or the same column, compute the quotient of the larger number divided by the smaller number, resulting in $n^2(n-1)$ fractions. The smallest of these fractions is called the "characteristic value" of this arrangement. Find the maximum value of the characteristic value.
Three, first prove: for any permutation $A$, its characteristic value
$$
l(A)=\frac{n+1}{n} \text {. }
$$

If among the $n$ natural numbers $n^{2}-n+1, n^{2}-n+2$, $\cdots, n^{2}$, there are two in the same row or column, then
$$
l(A) \leqslant \frac{a}{b} \leqslant \frac{n^{2}}{n^{2}-n+1}b$.
If all these $n$ large numbers are in different rows and columns, then $n^{2}-n$ must be in the same row and column as two of these numbers $a$ and $b$, respectively, and we have
$$
l(A) \leqslant \frac{a}{n^{2}-n} \leqslant \frac{n^{2}-1}{n^{2}-n}=\frac{n+1}{n},
$$

where $a$ is the smaller of $a$ and $b$.
For the permutation
$$
a_{i j}=\left\{\begin{array}{ll}
i+n(j-i-1), & i\frac{n+1}{n} .
\end{array}
$$

In the first column, we get an arithmetic sequence with a common difference of $d=n-1$
$$
\begin{array}{l}
n \leqslant(n-1)+n \leqslant(n-2)+2 n \leqslant \cdots \\
\leqslant 2+(n-2) n \leqslant 1+(n-1) n . \\
\text { Hence } \frac{a_{i 1}}{a_{k 1}}=\frac{1+i^{\prime} d}{1+k^{\prime} d} \geqslant \frac{1+n d}{1+(n-1) d} . \\
=\frac{n^{2}-n+1}{n^{2}-2 n+2} \geqslant \frac{n+1}{n} .
\end{array}
$$

When $n=2$, the last equality holds.
In columns $j=2,3, \cdots, n-1$, arranged from smallest to largest,
$$
\begin{array}{l}
j-1, j-2+n, j-3+2 n, \cdots, 1+(j-2) n, \\
n+(j-1) n, \cdots, j+(n-1) n,
\end{array}
$$

where the first $j-1$ terms form an arithmetic sequence with a common difference of $d=n-1$, and the last $n-j+1$ terms also form an arithmetic sequence with a common difference of $d=n-1$, and the difference between the $j$-th term and the $(j-1)$-th term is $2 n-1$. Thus,
$$
\frac{a_{i j}}{a_{k j}} \geqslant \frac{j+(n-1) n}{j+1+(n-2) n} \geqslant \frac{n+1}{n} .
$$

When $j=n-1$, the last equality holds.
In the $n$-th column, when $n \geqslant 3$,
$$
\frac{a_{\text {in }}}{a_{k n}} \geqslant \frac{n-1}{n-2}>\frac{n+1}{n} \text {. }
$$

Therefore, the maximum value of $l(A)$ is $\frac{n+1}{n}$.