[Solution] If we set $x=y-1, z=y+1$, then I becomes $3 y=46, y$ is not an integer, so $(A)$ is not true.

If we set $x=y-2, z=y+2$, then I becomes $3 y=46, y$ is not an integer, so $(B)$ is not true.
If we set $y=x+1, z=x+2, w=x+3$, then II becomes $4 x+6=46$, we get $x=10, y=11, z=12, w=13$. This is reasonable.
Therefore, the answer is $(C)$.
$31 \cdot 46$ If equation I: $x+y+z=46$, where $x, y, z$ are all positive integers. Equation II: $x+y+z+w=46$. Where $x, y, z, w$ are all positive integers, then
(A) I can be solved with consecutive integers.
(B) I can be solved with consecutive even numbers.
(C) II can be solved with consecutive integers.
(D) II can be solved with consecutive even numbers.
(E) I can be solved with consecutive odd numbers.
(11th American High School Mathematics Examination, 1960)