1. Find all arrays composed of four natural numbers $a, b, c, d$, such that the product of any three numbers divided by the remaining one leaves a remainder of 1. (Supplied by Li Shangzhi, University of Science and Technology of China)
Given that $bcd$ divided by $a$ leaves a remainder of 1, then $a \geqslant 2$. Otherwise, $bcd$ divided by 1 would leave a remainder of 0. Therefore, there exists an integer $k$ such that $bcd = k a - 1$, which means $a$ is coprime with $b, c, d$. Thus, $a, b, c, a'$ are all distinct and pairwise coprime.

Without loss of generality, we can rearrange the order so that $2 \leqslant ak$.

Since $a \geqslant 2$ and $k$ is a natural number, then $k=1$, and the natural number $a$ can be 2 or 3.

When $a=3$, using $ak \geqslant 1$ leads to a contradiction. Therefore, for $a$, there is only one possibility: $a=2$. Substituting $k=1$ and $a=2$ into equation (1), we get:
$$
\frac{1}{b} + \frac{1}{c} + \frac{1}{d} = \frac{1}{2} + \frac{1}{2bcd}.
$$

Since $b > \frac{1}{2}$, i.e., $b > a$, then $b$ has only two possibilities: $b=5$ or $b=3$.

When $b=5$, since $c, d$ are both coprime with 2 ($a=2$), from $b \frac{1}{b}, c < 12$. And $c$ is odd ($a=2, a, c$ are coprime), then $c \leqslant 11$. From (3), we have:
$$
d = 6 + \frac{35}{c-6}.
$$
Since $d \geqslant 7$, $\frac{35}{c-6}$ must be a natural number. Given $c \leqslant 11$, $c-6 \leqslant 5$, and of course $c-6 \geqslant 1$. Therefore, $c-6$ has only two possibilities: $c-6 = 1$ or $c-6 = 5$. When $c=7$, $d=41$; when $c=11$, $d=13$.

When $a < b < c < d$, it is easy to verify that $(a, b, c, d) = (2, 3, 7, 41)$ or $(2, 3, 11, 13)$ indeed satisfy the problem's requirements. Therefore, the solution set of the original problem is $\{a, b, c, d\} = \{2, 3, 7, 41\}$ or $\{a, b, c, d\} = \{2, 3, 11, 13\}$.