2. Arrange the $n^{2}$ natural numbers from 1 to $n^{2}$ randomly in an $n \times n$ square grid, where $n \geqslant 2$. For any pair of numbers in the same row or column, compute the ratio of the larger number to the smaller number. The minimum value among these $n^{2}(n-1)$ ratios is called the "characteristic value" of this arrangement. Find the maximum value of the "characteristic value".
First, we prove that for any arrangement $A$, its characteristic value
$$
c(A) \leqslant \frac{n+1}{n}.
$$

If the largest $n$ natural numbers $n^{2}-n+1, n^{2}-n+2, \cdots, n^{2}-1, n^{2}$ have two in the same row or column, then
$$
c(A) \leqslant \frac{a}{b} \leqslant \frac{n^{2}}{n^{2}-n+1}b.
$$
If these $n$ large numbers are in different rows and columns, when one of them is in the same row or column as $n^{2}-n$, we have
$$
c(A) \leqslant \frac{a}{n^{2}-n} \leqslant \frac{n^{2}-1}{n^{2}-n}=\frac{n+1}{n}.
$$

Here, $a$ is the smaller of the two largest numbers in the same row or column as $n^{2}-n$.
For the arrangement
$$
a_{i j}=\left\{\begin{array}{l}
i+n(j-i-1), \\
i+n(n-i+j-1),
\end{array} \text { when } i\frac{n+1}{n}.
\end{aligned}
$$

In the first column, we get an arithmetic sequence with a common difference of $d=n-1$
$$
\begin{aligned}
n & \leqslant(n-1)+n \leqslant(n-2)+2 n \leqslant \cdots \\
& \leqslant 2+(n-2) n \leqslant 1+(n-1) n.
\end{aligned}
$$

Thus, we have
$$
\begin{aligned}
\frac{a_{i 1}}{a_{k 1}} & =\frac{1+i d}{1+k d} \geqslant \frac{1+n d}{1+(n-1) d}=\frac{n^{2}-n+1}{n^{2}-2 n+2} \\
& \geqslant \frac{n+1}{n}.
\end{aligned}
$$

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

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 $d=n-1$. The difference between the $j$-th term and the $(j-1)$-th term is $2 n-1$. Thus, we have
$$
\frac{a_{i j}}{a_{k j}} \geqslant \frac{j+(n-1) n}{j+1+(n-2) n} \geqslant \frac{n+1}{n}.
$$

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