4. Let the 32 groups be $A_{1}, A_{2}, \cdots, A_{32}$, where the sum of the numbers in each group is 63, and such groups are called $A$-type groups; and the 63 groups be $B_{1}, B_{2}, \cdots, B_{63}$, where the sum of the numbers in each group is 32, and such groups are called $B$-type groups.

Clearly, each term $x_{k}$ belongs to exactly one $A$-type group and one $B$-type group, i.e., there are no common terms between groups of the same type; if two groups $A_{i}$ and $B_{j}$ have two common terms $x_{r}$ and $x_{t}$, then these two numbers can be merged into one term $x_{r}+x_{t}$, reducing the value of $n$, so we can assume that each pair of $A_{i}$ and $B_{j}$ has at most one common term.

Construct a graph model: use points $u_{1}, u_{2}, \cdots, u_{32}$ to represent $A_{1}, A_{2}, \cdots, A_{32}$, and points $v_{1}, v_{2}, \cdots, v_{63}$ to represent groups $B_{1}, B_{2}, \cdots, B_{63}$. If groups $A_{i}$ and $B_{j}$ have a common term, then connect a line between the corresponding points $u_{i}$ and $v_{j}$. Thus, we obtain a bipartite graph $G$, which has exactly $n$ edges and 95 vertices.
We will prove that $G$ is a connected graph.
If the largest connected component of graph $G$ is $G^{\prime}$, and the number of vertices in $G^{\prime}$ is less than 95, suppose in the connected component $G^{\prime}$, there are $a$ $A$-type vertices $u_{k_{1}}, u_{k_{2}}, \cdots, u_{k_{a}}$ and $b$ $B$-type vertices $v_{s_{1}}, v_{s_{2}}, \cdots, v_{s_{b}}$, where $a+b<95$. Then, in the corresponding $A$-type groups $A_{k_{1}}, A_{k_{2}}, \cdots, A_{k_{\varepsilon}}$ and $B$-type groups $B_{s_{1}}, B_{s_{2}}, \cdots, B_{s_{6}}$, any number $x_{i}$ in an $A$-type group $A_{r_{i}}$ appears in some $B$-type group $B_{s_{j}}$; and any number $x_{i}$ in a $B$-type group $B_{s_{i}}$ also appears in some $A$-type group $A_{r_{j}}$ (otherwise, there would be an edge connecting to a vertex outside the component, which is a contradiction).

Thus, the sum of the numbers in the $a$ $A$-type groups $A_{k_{1}}, A_{k_{2}}, \cdots, A_{k_{a}}$ should equal the sum of the numbers in the $b$ $B$-type groups $B_{s_{1}}, B_{s_{2}}, \cdots, B_{s_{6}}$, i.e., $63a = 32b$.
Therefore, $32|a$ and $63|b$.
So, $a+b \geqslant 32+63=95$, which is a contradiction.
Hence, $G$ is a connected graph, and graph $G$ has at least $95-1=94$ edges, i.e., $n \geqslant 94$.

On the other hand, we can actually construct a sequence of 94 terms $x_{1}, x_{2}, \cdots, x_{94}$ that satisfies the conditions of the problem. For example, take
\[
\begin{array}{l}
x_{1}=x_{2}=\cdots=x_{32}=32, \\
x_{33}=x_{34}=\cdots=x_{63}=31, \\
x_{64}=x_{65}=\cdots=x_{94}=1 .
\end{array}
\]

Thus, each $A$-type group can consist of one 32 and one 31, or one 32 and 31 ones, forming a total of 32 $A$-type groups, each with a sum of 63.

To obtain 63 $B$-type groups with a sum of 32, we can make $x_{1}, x_{2}, \cdots, x_{32}$ each form a group, and the remaining numbers can be combined into 31 $B$-type groups $\{1,31\}$.
In summary, the minimum value of $n$ is 94.
4. Given $n$ positive integers $x_{1}, x_{2}, \cdots, x_{n}$ whose sum is 2016. If these $n$ numbers can be divided into 32 groups with equal sums and also into 63 groups with equal sums, find the minimum value of $n$.