4. If the set $S=\{1,2,3, \cdots, 16\}$ is arbitrarily divided into $n$ subsets, then there must exist a subset in which there are elements $a, b, c$ (which can be the same), satisfying $a+b=c$. Find the maximum value of $n$.

[Note] If the subsets $A_{1}, A_{2}, \cdots, A_{n}$ of set $S$ satisfy the following conditions:
(1) $A_{i} \neq \varnothing(i=1,2, \cdots, n)$;
(2) $A_{i} \cap A_{j}=\varnothing$;
(3) $\bigcup_{i=1}^{n} A_{i}=S$,

then $A_{1}, A_{2}, \cdots, A_{n}$ is called a partition of set $S$.
4. First, when $n=3$, assume there exists a partition of the set that does not satisfy the condition. Thus, there must be a subset with at least six elements. Without loss of generality, let
$$
A=\left\{x_{1}, x_{2}, \cdots, x_{6}\right\}\left(x_{1}<x_{2}<\cdots<x_{6}\right) \text {. }
$$

Then $x_{6}-x_{1}, x_{6}-x_{2}, \cdots, x_{6}-x_{5} \notin A$. Among these, there must be three elements belonging to another subset, let
$$
x_{6}-x_{i}, x_{6}-x_{j}, x_{6}-x_{k} \in B(1 \leqslant i<j<k \leqslant 5) \text {. }
$$

Then $x_{j}-x_{i} 、 x_{k}-x_{j} 、 x_{k}-x_{i} \notin A \cup B$, but they cannot all belong to the third subset, leading to a contradiction.
Therefore, $n=3$ satisfies the condition.
Second, when $n=4$, the partition
$$
\{1\},\{2,3\},\{4,5,6,7,16\},\{8,9, \cdots, 15\}
$$

does not satisfy the condition.
In summary, $n_{\max }=3$.
[Note] \{1,4,7,10,13,16\}, \{2,3,8,9,14,15\}, \{5,6\}, \{11,12\} are also valid.