Three, 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 $x_{i, j}(i=1,2, \cdots, 100 ; j=1,2, \cdots, 25)$ (as shown in Table 1). Then, the numbers in each column of Table 1 are rearranged in descending order from top to bottom as $x^{\prime}{ }_{1, j} \geqslant x^{\prime}{ }_{2, j} \geqslant \cdots \geqslant$ $x^{\prime}{ }_{100, j}(j=1,2, \cdots, 25)$ (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 $\sum_{j=1}^{25} x^{\prime}{ }_{i, j} \leqslant 1$ 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$x^{\prime}{ }_{1,1}$ & $x^{\prime}{ }_{1,2}$ & $\cdots$ & $x^{\prime}{ }_{1,25}$ \\
\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}
Three, the minimum value of $k$ is 97.
(1) Take $x_{i, j}=\left\{\begin{array}{l}0,4(j-1)+1 \leqslant i \leqslant 4 j, \\ \frac{1}{24} \text {, for other } i .\end{array} \quad(j=1,2, \cdots, 25)\right.$

At this time, $\sum_{j=1}^{25} x_{i, j}=0+24 \times \frac{1}{24} .(i=1,2, \cdots, 100)$ satisfies the conditions of the problem. After rearrangement, we have
$$
x_{i, j}^{\prime}=\left\{\begin{array}{l}
\frac{1}{24}, 1 \leqslant i \leqslant 96, \\
0,97 \leqslant i \leqslant 100 .
\end{array}(j=1,2, \cdots, 25)\right.
$$

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

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

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

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