3i Let $n$ be a positive integer, and the indeterminate equation $x+2 y+2 z=n$ has exactly 28 sets of positive integer solutions. Find the value of $n$.


3. When $n$ is even, let $n=2 m$, then $x$ is even. For $x=2 k, 1 \leqslant k \leqslant m-1$, we have
$$y+z=m-k$$

In this case, there are $m-k-1$ solutions, meaning $x+2 y+2 z=2 m$ has a total of $(m-2)+(m-3)+\cdots+$
$$1+0=\frac{1}{2}(m-1)(m-2)$$

sets of positive integer solutions. Therefore,
$$\frac{1}{2}(m-1)(m-2)=28$$

we get
$$\begin{array}{l}
m=9 \\
n=18
\end{array}$$

Similarly, discussing the case $n=2 m+1$, we find $n=17$.
In summary, $n=17$ or 18.