$3 \cdot 89$ In a $100 \times 25$ rectangular grid, a non-negative real number is filled in each cell, with the number in the $i$-th row and $j$-th column being $x_{i j}(i=1,2, \cdots, 100 ; j=1,2, \cdots, 25)$ (as shown in Table 1). Then, the numbers in each column of Table 1 are rearranged in descending order from top to bottom as
$$
x^{\prime}{ }_{1 j} \geqslant x_{2 j} \geqslant \cdots \geqslant x^{\prime}{ }_{1000_{j}}, j=1,2, \cdots, 25,
$$
(see Table 2). Find the smallest natural number $k$, such that if 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.
[Solution] The minimum value of $k$ is 97.
Take
$$
x_{i j}=\left\{\begin{array}{l}
0, \text { when } 4(j-1)+1 \leqslant i \leqslant 4 j, \\
\frac{1}{24} \text {, otherwise, }
\end{array} \quad j=1,2, \cdots, 25 .\right.
$$

Then we have $\sum_{j=1}^{25} x_{i j}=0+\frac{1}{24} \times 24=1, i=1,2, \cdots, 100$. This shows that the set $\left\{x_{i j}\right\}$ satisfies the requirements of the problem. After rearrangement, we have
$$
x^{\prime}{ }_{i j}=\left\{\begin{array}{l}
\frac{1}{24}, \quad \text { when } 1 \leqslant i \leqslant 96, \\
0, \quad \text { when } 97 \leqslant i \leqslant 100,
\end{array} \quad j=1,2, \cdots, 25 .\right.
$$

At this point, $\sum_{j=1}^{25} x^{\prime}{ }_{i j}=25 \times \frac{1}{24}=\frac{25}{24}>1, i=1,2, \cdots, 96$. It is evident that the minimum value of $k$ is $\geqslant 97$.

On the other hand, the last 3 rows of Table 2 contain 75 numbers, which come from the 100 rows of Table 1. By the pigeonhole principle, there must be a row in Table 1 whose 25 numbers are all in the first 97 rows of Table 2. Thus, the 25 numbers in the 97th row of Table 2 are all no greater than the corresponding numbers in this row, so the sum of the numbers in the 97th row of Table 2 is no more than 1. This shows that the sum of the numbers in each of the last 4 rows is no more than 1. That is, $k=97$ meets the requirement.
In summary, the minimum value of $k$ is 97.