Example 8. Arrange $n^2$ distinct real numbers into the following table:
$$
\begin{array}{l}
a_{11}, a_{12}, \cdots, a_{1 n}, \\
a_{21}, a_{22}, \cdots, a_{2 n}, \\
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
a_{1_{1}}, a_{a_{2}}, \cdots, a_{n} .
\end{array}
$$

Take the maximum number from each row to get $n$ numbers, the smallest of which is $x$; then take the minimum number from each column to get $n$ numbers, the largest of which is $y$. Try to compare the size of $x$ and $y$.
Analysis: This problem is quite abstract and not easy to approach. It might be helpful to start by trying a few specific numbers to see what the relationship between $x$ and $y$ might be, which can guide us in finding a solution. Let's take 9 distinct real numbers and arrange them in the following table:
$$
\begin{array}{l}
1,5,7, \\
6,4,2, \\
8,10,9 .
\end{array}
$$

According to the problem, $x=6, y=4, x>y$. Now, let's rearrange the numbers as follows:
$$
\begin{array}{l}
6,5,7, \\
1,4,2, \\
10,8,9 .
\end{array}
$$

According to the problem, $x=4, y=4, x=y$. By trying a few more arrangements or using different sets of distinct real numbers, we can observe that no matter how we arrange them, we always have $x \geqslant y$. This is just an initial conjecture, and whether it holds in general still needs further analysis. From the table, we can see that when $x$ is in a certain row, $x$ is not less than the first number in that row; when $y$ is in a certain column, $y$ is not greater than the first number in that column. If $b$ is the number at the intersection of the row where $x$ is located and the column where $y$ is located, then we must have $x \geqslant b, y \leqslant b \Rightarrow x \geqslant y$.

Therefore, by using the number at the intersection of the row where $x$ is located and the column where $y$ is located as an auxiliary element, we can determine the relationship between $x$ and $y$, leading to a simpler solution.

Solution: Let $x$ be in the $i$-th row ($1 \leqslant i \leqslant n$) and $y$ be in the $j$-th column ($1 \leqslant j \leqslant n$). According to the problem, we have $x \geqslant a_{1}, y \leqslant a_{1} \Rightarrow x \geqslant y$.
1) When $n$ is odd, $x^{2} \geqslant y^{2}$;
2) When $n$ is even,
(1) If $x \geqslant y \geqslant 0$, then $x^{2} \geqslant y^{2}$;
(2) If $0 \geqslant x \geqslant y$, then $x^{2} \leqslant y^{2}$;
(3) If $x \geqslant 0 \geqslant y$, then when $|x| \geqslant |y|$, $x^{2} \geqslant y^{2}$; when $|x| \leqslant |y|$, $x^{2} \leqslant y^{2}$.