6. 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 $x_{1, j}^{\prime} \geqslant x_{2, j}^{\prime} \geqslant \cdots \geqslant x_{100, j}^{\prime}(j=1,2, \cdots, 25)$, as shown in 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 $\sum_{j=1}^{25} x_{i, j}^{\prime} \leqslant 1$ 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$x^{\prime}{ }_{1,1}$ & $x_{1,2}^{\prime}$ & $\cdots$ & $x^{\prime}{ }_{1,25}$ \\
\hline$x^{\prime}{ }_{2,1}$ & $x^{\prime}{ }_{2,2}$ & $\cdots$ & $x^{\prime}{ }_{2,25}$ \\
\hline$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
\hline$x_{100,1}^{\prime}$ & $x^{\prime}{ }_{100,2}$ & $\cdots$ & $x_{100,25}^{\prime}$ \\
\hline
\end{tabular}
6. Let $x_{1,1}=x_{2,1}=x_{3,1}=x_{4,1}=x_{5,2}=x_{6,2}=$ $x_{7,2}=x_{8,2}=0, \cdots, x_{97,25}=x_{98,25}=x_{99,25}=x_{100,25}$ $=0$, the rest of the elements are $\frac{1}{24}$, then the sum of 25 numbers in each row is $0+$ $24 \times \frac{1}{24}=1$, at this point in Table 1, each column has exactly 4 zeros, so after adjusting to Table 2, the last four rows are all 0, but the numbers in the first 96 rows are all $\frac{1}{24}$, the sum of each row is $\frac{25}{24}>1$. This shows that the smallest $K \geqslant 97$. The following proves that the minimum value of $K$ is 97. Because the last three rows (the 98th, 99th, and 100th rows) can only accommodate 75 elements, there must be a row (let's say the $r$-th row) in Table 1, all of its 25 elements, in the adjusted Table 2, are in the first 97 rows. Otherwise, each row has at least one element falling into the last three rows, then at least 100 elements fall into the last three rows, which is impossible. This means that in Table 2, $x^{\prime}{ }_{97, j} \leqslant x_{r, j}\left(x_{r, j}\right.$ is still in the $j$-th column of Table 2, the row number is not necessarily $r$. But the row number $\leqslant$ 97). Therefore, for any $i \geqslant 97$, $x^{\prime}{ }_{i, j} \leqslant x^{\prime}{ }_{47, j} \leqslant x_{r, j}$. So when $i \geqslant 97$, $\sum_{j=1}^{25} x^{\prime}{ }_{i, j} \leqslant \sum_{j=1}^{25} x_{r, j} \leqslant 1$. Hence, the minimum value of $K$ is 97.