13.24 A pile of stones has a total weight of 100 kilograms, where the weight of each stone does not exceed 2 kilograms. By taking out some of these stones in various ways, and finding the difference between the sum of the weights of these stones and 10 kilograms, let the minimum value of the absolute values of these differences be \(d\), find the maximum value of \(d\).
[Solution] To make the value of $d$ as large as possible, the weight of each stone should be as large as possible, of course, not exceeding 2 kilograms. However, when the weight is too close to 2 kilograms, the difference between the total weight of 5 stones and 10 kilograms becomes smaller. For this, if we set the weight of the stone as $x$, we can have
$$
10-5 x=6 x-10 \text {. }
$$

Solving for $x$ gives $x=\frac{20}{11}$. It is easy to see that when taking 55 stones, each weighing $\frac{20}{11}$ kilograms, the total weight is 100 kilograms, and $d=\frac{10}{11}$. Therefore, the maximum value of $d$ is no less than $\frac{10}{11}$.

On the other hand, if the maximum value of $d$ is $D$ and $D$ is greater than $\frac{10}{11}$, then there must exist a pile of stones with a total weight of 100 kilograms, and the weight $M$ of any group of stones taken from it must satisfy the inequality
$$
|M-10|>\frac{10}{11} \text {. }
$$

We examine such a group of stones, whose total weight $M>10$, but removing any one of them results in a total weight less than 10. Let the weights of these stones be $x_{1}, x_{2}, \cdots, x_{k}$. Then from (1) we have
$$
x_{1}+x_{2}+\cdots+x_{k}=M>10+\frac{10}{11} \text {. }
$$

And for each $x_{i}$, we have
$$
M-x_{i}<10-\frac{10}{11}, i=1,2, \cdots, k .
$$

Since $x_{i} \leqslant 2, i=1,2, \cdots, k$, we have
$$
10 \geqslant x_{1}+x_{2}+x_{3}+x_{4}+x_{5}>5 \times \frac{10}{11} .
$$

This indicates that the difference between the total weight of 5 stones and 10 kilograms is less than $\frac{10}{11}$, which contradicts (1).
In summary, the maximum value of $d$ is $\frac{10}{11}$.