4. Find the largest positive integer $k$ such that the set of positive integers can be partitioned into $k$ subsets $A_{1}, A_{2}, \cdots, A_{k}$, so that for all integers $n(n \geqslant 15)$ and all $i \in\{1,2, \cdots, k\}$, there exist two distinct elements in $A_{i}$ whose sum is $n$.
4. The largest positive integer $k$ is 3.

When $k=3$, let
$A_{1}=\{1,2,3\} \cup\{3 m \mid m \in \mathbf{Z}$, and $m \geqslant 4\}$,
$A_{2}=\{4,5,6\} \cup\{3 m-1 \mid m \in \mathbf{Z}$, and $m \geqslant 4\}$,
$A_{3}=\{7,8,9\} \cup\{3 m-21 m \in \mathbf{Z}$, and $m \geqslant 4\}$.
Then the sum of two different elements in $A_{1}$ can represent all
positive integers $n$ greater than or equal to $1+12=13$; the sum of two different elements in $A_{2}$ can represent all positive integers $n$ greater than or equal to $4+11=15$; the sum of two different elements in $A_{3}$ can represent all positive integers $n$ greater than or equal to $7+10=17$ and $7+8=15, 7+9=16$.

If $k \geqslant 4$, and there exist sets $A_{1}, A_{2}, \cdots, A_{k}$ that satisfy the conditions, then the sets $A_{1}, A_{2}, A_{3}, A_{4} \cup A_{5} \cup \cdots \cup A_{k}$ also satisfy the conditions.
Thus, we can assume $k=4$.
Let $B_{i}=A_{i} \cap\{1,2, \cdots, 23\}(i=1,2,3,4)$.
For each $i=1,2,3,4$, the integers $15,16, \cdots, 24$ can be written as the sum of two different elements in $B_{i}$. Therefore, $B_{i}$ must have at least five elements.

Since $\left|B_{1}\right|+\left|B_{2}\right|+\left|B_{3}\right|+\left|B_{4}\right|=23$, there exists $j \in\{1,2,3,4\}$ such that $\left|B_{j}\right|=5$.

Let $B_{j}=\left\{x_{1}, x_{2}, \cdots, x_{5}\right\}$. Then the sums of two different elements in $A_{j}$ that represent the integers $15,16, \cdots, 24$ are exactly the sums of all two different elements in $B_{j}$.
$$
\begin{array}{l}
\text { Hence } 4\left(x_{1}+x_{2}+\cdots+x_{5}\right) \\
=15+16+\cdots+24=195 .
\end{array}
$$

Since 195 is not divisible by 4, the above equation does not hold.
Thus, a contradiction arises.