［Solution］The given equation can be transformed into $2(x+y)+z=n$.
Also, since $x, y, z$ are all positive integers, $x+y$ can take values $2,3,4, \cdots,\left[\frac{n}{2}\right]$ (when $n$ is odd) or $\left[\frac{n}{2}\right]-1$ (when $n$ is even).

For each $x+y=k$, the pair $(x, y)$ can have $k-1$ different forms $(1, k-1), \quad(2, k-2), \cdots, \quad(k-2,2), \quad(k-1,1)$.
Notice that $1+2+3+4+5+6+7=28$,
thus $x+y=8$, i.e., $\left[\frac{n}{2}\right]=8$ or $\left[\frac{n}{2}\right]-1=8$,
so $n=17$ or 18.
Therefore, the answer is $(D)$.
$31 \cdot 13 n$ is a positive integer. The equation $2 x+2 y+z=n$ has 28 sets of positive integer solutions for $x, y, z$, then $n$ is
(A) 14 or 15.
(B) 15 or 16.
(C) 16 or 17.
(D) 17 or 18.
(E) 18 or 19.
(40th American High School Mathematics Examination, 1989)