Example 13 In a $100 \times 25$ rectangular table, each cell is filled with a non-negative real number, the number filled 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 descending order from top to bottom as $x^{\prime}_{1j} \geqslant x^{\prime}_{2j} \geqslant \cdots \geqslant x^{\prime}_{100j} (j=1,2, \cdots, 25)$; this is recorded as Table 2.

Find the smallest positive integer $k$, such that if the numbers filled in Table 1 satisfy $\sum_{j=1}^{25} x_{ij} \leqslant 1 (i=1,2, \cdots, 100)$, then when $i \geqslant k$, in Table 2 it can be guaranteed that: $\sum_{j=1}^{25} x^{\prime}_{ij} \leqslant 1$.
(1997 National High School League Additional Test Question)
\begin{tabular}{|c|c|c|c|}
\hline$x_{1,1}$ & $x_{1,2}$ & $\cdots$ & $x_{1,25}$ \\
\hline$x_{2,1}$ & $x_{2,2}$ & $\cdots$ & $x_{2,25}$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline$x_{100,1}$ & $x_{100,2}$ & $\cdots$ & $x_{100,25}$ \\
\hline
\end{tabular}
\begin{tabular}{|c|c|c|c|}
\hline$x^{\prime}_{1,1}$ & $x^{\prime}_{1,2}$ & $\cdots$ & $x^{\prime}_{1,25}$ \\
\hline$x^{\prime}_{2,1}$ & $x^{\prime}_{2,2}$ & $\cdots$ & $x^{\prime}_{2,25}$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline$x^{\prime}_{100,1}$ & $x^{\prime}_{100,2}$ & $\cdots$ & $x^{\prime}_{100,25}$ \\
\hline
\end{tabular}

Table 1
Table 2
If in Table 1 each row has at least one number that does not appear in the first $k$ rows of Table 2, then the first $k$ rows of Table 2 contain at most $24 \times 100 = 2400$ numbers from Table 1, hence $25 k \leqslant 2400$, which means $k \leqslant 96$.

Therefore, when $k \geqslant 97$, there must be a row in Table 1 (let's say the $r$-th row) whose 25 numbers all appear in the first $k$ rows of Table 2. Thus, by the construction rule of Table 2, when $i \geqslant k$, we have $x_{i j}^{\prime} \leqslant x_{r j} (j=1,2, \cdots, 25)$, and by the given condition,
$$
\sum_{j=1}^{25} x^{\prime}{ }_{i j} \leqslant \sum_{j=1}^{25} x_{i j} \leqslant 1 .
$$

Therefore, $k_{\min } \leqslant 97$.
On the other hand, take
$$
x_{i j}=\left\{\begin{array}{l}
0(4(j-1)+1 \leqslant i \leqslant 4 j) \\
\left.\frac{1}{24} \text { (the rest of } i\right)
\end{array} \quad(j=1,2, \cdots, 25) .\right.
$$

At this time, $\sum_{j=1}^{25} x_{i j}=24 \times \frac{1}{24}+0=1(i=1,2, \cdots, 100)$ satisfies the condition, and after rearrangement, in Table 2 we have
$$
x^{\prime}{ }_{i j}=\left\{\begin{array}{l}
\frac{1}{24}(1 \leqslant i \leqslant 96) \\
0(97 \leqslant i \leqslant 100)
\end{array}(j=1,2, \cdots, 25) .\right.
$$

At this time, $\sum_{j=1}^{25} x_{i j}=\frac{1}{24} \times 25=\frac{25}{24}>1(1 \leqslant i \leqslant 96)$.
Hence, $k_{\text {min }} \geqslant 97$.
In summary, the minimum value of $k$ is 97.