1.94 Find all quadruples of positive integers $(a, b, c, d)$ such that the product of any three of them, when divided by the remaining one, leaves a remainder of 1.
(China National Team Selection Test, 1994)
[Solution] First, we prove that $a, b, c, d$ are all no less than 2 and are pairwise coprime.
This is because, according to the problem, we have $b c d = k a + 1$. Clearly, $a \geqslant 2$, otherwise, if $a = 1$, then $b c d$ would be divisible by $a$, which contradicts the problem statement.
At the same time, $a$ is coprime with any of $b, c, d$.
Similarly, we have $b \geqslant 2, c \geqslant 2, d \geqslant 2$, and $a, b, c, d$ are pairwise coprime.
Without loss of generality, assume $2 \leqslant a \leqslant b \leqslant c \leqslant d$.
Equation (1) can be rewritten as
$$
1 + \frac{1}{6 c d} = \frac{1}{2} + \frac{1}{3} + \frac{1}{c} + \frac{1}{d},
$$

Thus, $\frac{1}{c} + \frac{1}{d} = \frac{1}{6} + \frac{1}{6 c d} \approx \frac{1}{6}$, and $c < 12$.
Given $a = 2$, we know $c$ is an odd number, so $c \leqslant 11$.
From equation (3), we can derive $d = 6 + \frac{35}{c - 6}$.
Since $d \geqslant 7$ and $d$ is a positive integer, $\frac{35}{c - 6}$ must be a positive integer. Considering $c \leqslant 11$, we have $1 \leqslant c - 6 \leqslant 5$, so $c - 6 = 1$ or $c - 6 = 5$.
Thus, $c = 7$ or $c = 11$.
When $c = 7$, from equation (4), $d = 41$,
When $c = 11$, from equation (4), $d = 13$.
Therefore, the four numbers $(a, b, c, d) = (2, 3, 7, 11)$ or $(2, 3, 11, 13)$.