9.48 The total weight of a pile of stones is 100 kilograms, where the weight of each stone does not exceed 2 kilograms. By taking out some of the stones in various ways and calculating the difference between the sum of the weights of these stones and 10 kilograms. Among all these differences, the minimum value of their absolute values is denoted as $d$. Among all piles of stones that meet the above conditions, find the maximum value of $d$.
[Solution] Take any pile of stones that satisfies the conditions of the problem, and let their weights be $x_{1}, x_{2}$, $\cdots, x_{n}$, without loss of generality, assume $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{n}$. According to the definition of $d$, there exists a natural number $k$ such that
$$
x_{1}+x_{2}+\cdots+x_{k} \geqslant 10+d \text { and } x_{1}+x_{2}+\cdots+x_{k-1} \leqslant 10-d \text {. }
$$

From this, we can deduce that $x_{1} \geqslant x_{2} \geqslant \cdots \geqslant x_{k} \geqslant 2 d$. Also, $05$. Therefore, we have
$$
10 d \leqslant x_{1}+x_{2}+x_{3}+x_{4}+x_{5} \leqslant 10-d,
$$

which implies $d \leqslant \frac{10}{11}$.
Furthermore, take 55 stones each weighing $\frac{10}{11}$ kilograms. It is easy to see that in this case $d=\frac{10}{11}$. Therefore, the maximum value of $d$ is $\frac{10}{11}$.