Example 8 In a $100 \times 25$ rectangular table, each cell is filled with a non-negative real number, the number in the $i$-th row and $j$-th column is denoted as $x_{ij}$, as shown in Table 1. Then, the numbers in each column of Table 1 are rearranged in non-increasing order as $x_{1j}^{\prime} \geqslant x_{2j}^{\prime} \geqslant \cdots \geqslant x_{100j}^{\prime}$, forming Table 2.

Find the smallest natural number $k$, such that if the numbers filled in Table 1 satisfy $\sum_{j=1}^{25} x_{ij} \leqslant 1$ for $i=1,2, \cdots, 100$, then when $i \geqslant k$, in Table 2, it is guaranteed that: $\sum_{j=1}^{25} x_{ij}^{\prime} \leqslant 1$ holds.
(1997, National High School Mathematics Competition)
Solution: First, consider a row $x_{r 1}, x_{r 2}$, $\cdots, x_{r 25}$ in Table 1, which must appear in the first few items of Table 2, because $100 \times 24 = 2400$, while $97 \times 25 = 2425$. Therefore, a row in Table 1 must appear in the first 97 items of Table 2.

Thus, when $i \geqslant 97$, $x_{i j}^{\prime} \leqslant x_{97 j}^{\prime} \leqslant x_{i j} (j=1$, $2, \cdots, 25)$,
so when $i \geqslant 97$, $\sum_{j=1}^{25} x_{i j}^{\prime} \leqslant \sum_{j=1}^{25} x_{r j} \leqslant 1$.
On the other hand, take
$$
x_{i j}=\left\{\begin{array}{ll}
0 & (4(j-1)+1 \leqslant i \leqslant 4 j), \\
\frac{1}{24} & (\text { otherwise } i) .
\end{array}\right.
$$
$(j=1,2, \cdots, 25)$, at this time $\sum_{j=1}^{25} x_{i j}=1(i=1,2$, $\cdots, 100)$,
after rearrangement
$$
\begin{array}{l}
x_{i j}^{\prime}=\left\{\begin{array}{ll}
\frac{1}{24} & (1 \leqslant i \leqslant 96, j=1,2, \cdots, 25), \\
0 & (97 \leqslant i \leqslant 100) .
\end{array}\right. \\
\text { we have } \sum_{j=1}^{25} x_{i j}^{\prime}=25 \times \frac{1}{24}>1(1 \leqslant i \leqslant 96) .
\end{array}
$$

Therefore, $k \geqslant 97$, i.e., the minimum value of $k$ is 97.
5 Comprehensive Estimation
To solve discrete extremum problems, it often requires the integrated use of various strategic ideas for extremum estimation.