Four, (50 points) Given that $n$ is an integer and $n \geqslant 2$, the elements of set $A$ are all positive integers, the smallest element of $A$ is 1, and the largest element is $a$, and $7 \times 3^{n}<a<3^{n+2}$, satisfying that for any element $x(x \neq 1)$ in $A$, there exist $s$, $t$, $p \in A$ such that $x=s+t+p$. Find the minimum number of elements in set $A$.
The minimum value sought is $n+4$.
On the one hand, when $|A|=n+4$, take
$$
A=\left\{1,3,3^{2}, \cdots, 3^{n+1}, 7 \times 3^{n}, 7 \times 3^{n}+2\right\} \text {. }
$$

Then $7 \times 3^{n}=3^{n}+2 \times 3^{n+1}$,
$7 \times 3^{n}+2=7 \times 3^{n}+2 \times 1$,
$3^{k+1}=3^{k}+3^{k}+3^{k}(0 \leqslant k \leqslant n)$.
On the other hand, let $A=\left\{a_{1}, a_{2}, \cdots, a_{m}\right\}$, where $a_{1}=1<a_{2}<\cdots<a_{m}=a, a_{i} \in \mathbf{Z}(i=1,2, \cdots, m)$.

By the problem statement, for any $2 \leqslant i \leqslant m(i \in \mathbf{N})$ there exist $1 \leqslant j, k, h \leqslant i-1$, such that
$a_{i}=a_{j}+a_{k}+a_{h} \leqslant a_{i-1}+a_{i-1}+a_{i-1}=3 a_{i-1}$.
Then $7 \times 3^{n}<a_{m}=a \leqslant 3^{m-1} a_{1}=3^{m-1}$
$\Rightarrow m-1 \geqslant n+2 \Rightarrow m \geqslant n+3$.
Assume $m=n+3$.
Prove by mathematical induction that $a_{i}=3^{i-1}(1 \leqslant i \leqslant m)$.
Clearly, $a_{2}=3 a_{1}=3$, so the conclusion holds for $i=1,2$;
Assume that for $1 \leqslant i \leqslant t$, $a_{i}=3^{i-1}$.
When $i=t+1$, if $a_{t+1} \neq 3^{t}=3 a_{t}$, then $a_{t+1} \leqslant 2 a_{t}+a_{t-1}=7 \times 3^{t-2}$.
But from equation (1) we know
$$
a_{m} \leqslant 3^{m-1-t} a_{t+1} \leqslant 3^{m-1-t} \times 7 \times 3^{t-2}=7 \times 3^{n} \text {, }
$$

which contradicts the problem statement.
Therefore, $a_{t+1}=3^{t}$. At this point,
$$
A=\left\{1,3,3^{2}, \cdots, 3^{n+1}, a\right\}, a=3^{n+2},
$$

which contradicts the condition.
Thus, when $m=n+3$, it is impossible.
In summary, the minimum number of elements in $A$ is $n+4$.