Problem 1368

[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 $\frac{10}{11}$.