Three. (This question is worth 50 points) 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 $x_{i, j} (i=1,2, \cdots, 100 ; j=1,2, \cdots, 25)$ (as shown in Table 1). Then, the numbers in each column of Table 1 are rearranged in descending order from top to bottom as
$$
\begin{array}{c}
x_{1, j}^{\prime} \geqslant x_{2, j}^{\prime} \geqslant \cdots \geqslant x_{100, j}^{\prime} \\
(j=1,2, \cdots, 25) . \text { (as shown in Table 2) }
\end{array}
$$

Find the smallest natural number $k$, such that if the numbers 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
$$
\sum_{j=1}^{25} x_{i, j}^{\prime} \leqslant 1
$$

holds.
Table 1
\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 2
\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}
(Proposed by the Problem Committee)
Three, the minimum value of $k$ is 97.
(1) 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, $\sum_{j=1}^{25} x_{i, j}=0+24 \times \frac{1}{24} .(i=1,2, \cdots, 100)$

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

At this time, $\sum_{j=1}^{25} x_{i, j}^{\prime}=25 \times \frac{1}{24}>1 .(1 \leqslant i \leqslant 96)$
Therefore, the minimum value of $k$ $\geqslant 97$.
(2) First, we prove: In Table 1, there must be a row (let's say the $r$-th row) with all the numbers
$$
x_{r, 1}, x_{r, 2}, \cdots, x_{r, 25}
$$

which must appear in the first 97 rows of the rearranged Table 2.
In fact, if the above conclusion does not hold, then in Table 1, each row must have at least one number that does not appear in the first 97 rows of Table 2, meaning that the first 97 rows of Table 2 contain at most $100 \times 24=2400$ numbers from Table 1. This contradicts the fact that the first 97 rows of Table 2 contain $25 \times 97=2425$ numbers.

Secondly, by the rearrangement requirement, the numbers in each column of Table 2 are arranged from top to bottom in descending order, so when $i \geqslant 97$,
$$
x_{i, j}^{\prime} \leqslant x_{97, j}^{\prime} \leqslant x_{r, j} .(j=1,2, \cdots, 25)
$$

Therefore, when $i \geqslant 97$,
$$
\sum_{j=1}^{25} x_{i, j}^{\prime} \leqslant \sum_{j=1}^{25} x_{r, j} \leqslant 1 .
$$

Combining (1) and (2), the minimum value of $k$ is 97.
(Liu Yujie provided)