The largest positive integer $k$ is 3.
When $k=3$, let
$$
\begin{array}{l}
A_{1}=\{1,2,3\} \cup\{12,15,18, \cdots\}, \\
A_{2}=\{4,5,6\} \cup\{11,14,17, \cdots\}, \\
A_{3}=\{7,8,9\} \cup\{10,13,16, \cdots\} .
\end{array}
$$

Then any positive integer $n$ not less than $1+12=13$ can be expressed as the sum of two distinct elements in $A_{1}$;

any positive integer $n$ not less than $4+11=15$ can be expressed as the sum of two distinct elements in $A_{2}$;

any positive integer $n$ not less than $7+10=17$ and $7+8=15, 7+9=16$ can be expressed as the sum of two distinct elements in $A_{3}$.

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, reducing to the case where $k=4$.
Thus, we can assume $k=4$,
$$
B_{i}=A_{i} \cap\{1,2, \cdots, 23\}(i=1,2,3,4) .
$$

Since $\left|B_{1}\right|+\left|B_{2}\right|+\left|B_{3}\right|+\left|B_{4}\right|=23$, we can assume without loss of generality that $\left|B_{1}\right|$ is the smallest. Then $\left|B_{1}\right| \leqslant 5$, and the integers $15,16, \cdots, 24$ can be written as the sum of two distinct elements in $B_{1}$.

Thus, $\left|B_{1}\right|=5$, and the integers $15,16, \cdots, 24$ are the ten possible sums of two distinct elements in $B_{1}$.
Hence $4\left(x_{1}+x_{2}+\cdots+x_{5}\right)$
$$
=15+16+\cdots+24=195 \text {. }
$$

Since 195 is not divisible by 4, this leads to a contradiction.
Example 5 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 $n .^{[4]}$