9.48 A pile of stones has a total weight of 100 kilograms, with each stone weighing no more than 2 kilograms. By taking some of these stones in various ways and calculating the difference between the sum of their weights and 10 kilograms. Among all these differences, the minimum absolute value 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
$$x_{1}+x_{2}+\cdots+x_{k} \geqslant 10+d \quad \text { and } x_{1}+x_{2}+\cdots+x_{k-1} \leqslant 10-d \text {. }$$

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, taking 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 $\frac{10}{11}$.