1 Place 1, 2, 3, 4, 2, 6, 7, 8 at the eight vertices of a cube: so that the sum of any four numbers on the same face is not less than 10. Find the minimum value of the sum of the four numbers on each face. (Provided by Dai Zonghu)
Let the four numbers $a_{1}, a_{2}, a_{3}, a_{4}$ on a certain face reach the minimum sum,
A. $a_{1}<a_{2}<a_{3}<a_{4}$. Since the sum of three different positive integers less than 5 is at most 9, hence $a_{i} \geq 6$. Therefore,
$$
a_{1}+a_{2}+a_{3}+a_{4} \geqslant 16 .
$$

As shown in the figure, we take the 12 edges of the cube to be $1, 7$, 
3, 6, and the four faces to be $4, 8, 2, 5$, then the sum of the side faces is $6+3+$ 
$2+5=16$. This shows that 16 is achievable.