$2 \cdot 77$ 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 $x_{i j}(i=1,2, \cdots, 100 ; j=1,2, \cdots, 25)$ (as shown in Table 1). However, the numbers in each column of Table 1 are rearranged in descending order from top to bottom as $x_{1 . j}^{\prime} \geqslant x_{2 j}^{\prime} \geqslant \cdots \geqslant x_{100 j}^{\prime}(j=1,2, \cdots, 25)$
\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}

Table 1
\begin{tabular}{|c|c|c|c|}
\hline$x_{1,1}^{\prime}$ & $x_{1,2}^{\prime}$ & $\cdots$ & $x_{1,25}^{\prime}$ \\
\hline$x_{2,1}^{\prime}$ & $x_{2,2}^{\prime}$ & $\cdots$ & $x_{2,25}^{\prime}$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline$x_{100,1}^{\prime}$ & $x_{100,2}^{\prime}$ & $\cdots$ & $x_{100,25}^{\prime}$ \\
\hline
\end{tabular}

Table 2 (as shown in Table 2) Find the smallest natural number $k$, such that as long as the numbers filled in Table 1 satisfy $\sum_{j=1}^{25} x_{i j} \leqslant 1$, $(i=1,2, \cdots, 100)$, then when $i \geqslant k$, in Table 2 it can be guaranteed that
$$\sum_{j=1}^{25} x_{i j} \leqslant 1$$

holds.
[Solution] The minimum value of $k$ is 97.
In fact, if we take
$$x_{i j}=\left\{\begin{array}{l}
0,4,(j-1)+1 \leqslant i \leqslant 4 j \\
\frac{1}{24} . \text { for the rest of } i
\end{array} \quad(j=1,2, \cdots, 25)\right.$$

Then, we have
$$\sum_{j=1}^{25} x_{i j}=0+24 \times \frac{1}{24}=1 \quad(i=1,2, \cdots, 100)$$

which satisfies the problem's conditions. After rearrangement, we have
$$x_{i j}^{\prime}=\left\{\begin{array}{l}
\frac{1}{24}, 1 \leqslant i \leqslant 96, \\
0, \quad 97 \leqslant i \leqslant 100 .
\end{array} \quad(j=1,2, \cdots, 25)\right.$$

At this point,
$$\sum_{j=1}^{25} x_{i j}^{\prime}=25 \times \frac{1}{24}>1, \quad(1 \leqslant i \leqslant 96)$$

Therefore, the minimum value of $k$ $\geqslant 97$.
On the other hand, since the last three rows of Table 2 contain only 75 numbers, and Table 1 has 100 rows, there must be at least one row in Table 1 where all its elements appear in the first 97 rows of Table 2. Suppose the $r$-th row of Table 1 has all its numbers
$$x_{r, 1}, x_{r, 2}, \cdots, x_{r .25}$$

appearing in the first 97 rows of Table 2. Thus, when $i \geqslant 97$, we have
$$\begin{array}{l}
x_{i j}^{\prime} \leqslant x_{r j}, \quad(j=1,2, \cdots, 25) \\
\sum_{j=1}^{25} x_{i j}^{\prime} \leqslant \sum_{j=1}^{25} x_{r j} \leqslant 1
\end{array}$$

In summary, the minimum value of $k$ is 97.