Example $\mathbf{3}$ (2003 Western China Mathematical Olympiad) Place the numbers 1, 2, 3, 4, 5, 6, 7, 8 at the eight vertices of a cube so that the sum of any three numbers on each face is not less than 10. Find the minimum value of the sum of the four numbers on each face.
Let's assume that the sum of four numbers $a_{1}, a_{2}, a_{3}, a_{4}$ on a certain face reaches the minimum value, and $a_{1}<a_{2}<a_{3}<a_{4}$. Since the sum of three different positive integers less than 5 is at most 9, it follows that $a_{4} \geqslant 6$, thus $a_{1}+a_{2}+a_{3}+a_{4} \geqslant 16$.
As shown in Figure 1-1, we take the top face of the cube to be $1, 7, 3, 6$ in sequence, and the bottom face to be $4, 8, 2, 5$ in sequence, then the right side face is $6+3+2+5=16$, which indicates that 16 is achievable.