Four, let $n$ sets $A_{1}, A_{2}, \cdots, A_{n}$ be a partition of the set $A=\{1,2, \cdots, 29\}$, and the sum of any elements in $A_{i}(i=1,2$, $\cdots, n)$ does not equal 30. Find the minimum possible value of $n$.

[Note] If the non-empty subsets $A_{1}, A_{2}, \cdots, A_{n}$ $\left(n \in \mathbf{N}_{+}, n \geqslant 2\right)$ of set $A$ satisfy
$$
\begin{array}{l}
A_{i} \cap A_{j}=\varnothing(i \neq j), \\
A_{1} \cup A_{2} \cup \cdots \cup A_{n}=A,
\end{array}
$$

then $A_{1}, A_{2}, \cdots, A_{n}$ is called a partition of set $A$.
The minimum value of $n$ is 3.
First, we decompose
$$
\begin{array}{l}
A_{1}=\{1,2, \cdots, 7\}, \\
A_{2}=\{10,11, \cdots, 15,21,22\}, \\
A_{3}=\{8,9,16,17, \cdots, 20,23,24, \cdots, 29\}
\end{array}
$$

which satisfies the conditions.
Next, we prove that $n=2$ does not satisfy the conditions.
Assume, for contradiction, that $A$ can be decomposed into two sets $A_{1}$ and $A_{2}$, and without loss of generality, let $1 \in A_{1}$, and the smallest element in $A_{1}$ other than 1 be $k$. Clearly, $k$ exists.
(1) If $k \geqslant 9$, then $2,3,4,6,7,8 \in A_{2}$, and $2+3+4+6+7+8=30$, which is a contradiction.
(2) If $2 \leqslant k \leqslant 8$, then consider $k+1$.
If $k+1 \in A_{2}$, then $29-k > k+1$, and no matter which set the element $29-k$ belongs to, it leads to a contradiction. Therefore, $k+1 \in A_{1}$.

Similarly, $k+2 \in A_{1}, k+3 \in A_{1}, \cdots, 11 \in A_{1}$, then $9, 10, 11 \in A_{1}$. But $9+10+11=30$, which is a contradiction.
Thus, $n=2$ does not satisfy the conditions.
In conclusion, the minimum value of $n$ is 3.