Isabel wants to partition the set $\mathbb{N}$ of the positive integers into $n$ disjoint sets $A_{1}, A_{2}, \ldots, A_{n}$. Suppose that for each $i$ with $1\leq i\leq n$, given any positive integers $r, s\in A_{i}$ with $r\neq s$, we have $r+s\in A_{i}$. If $|A_{j}|=1$ for some $j$, find the greatest positive integer that may belong to $A_{j}$.
1. **Understanding the Problem:**
   We need to partition the set of positive integers $\mathbb{N}$ into $n$ disjoint sets $A_1, A_2, \ldots, A_n$ such that for each $i$ with $1 \leq i \leq n$, given any positive integers $r, s \in A_i$ with $r \neq s$, we have $r + s \in A_i$. Additionally, we know that $|A_j| = 1$ for some $j$. We need to find the greatest positive integer that may belong to $A_j$.

2. **Analyzing the Condition:**
   The condition $r + s \in A_i$ for $r, s \in A_i$ implies that each $A_i$ must be closed under addition. This means that if $A_i$ contains any two distinct elements, it must contain all their sums, and thus all multiples of their greatest common divisor (gcd).

3. **Implication of Singleton Set:**
   Since $|A_j| = 1$ for some $j$, let $A_j = \{k\}$. This means $k$ is the only element in $A_j$ and no other element can be in $A_j$.

4. **Partitioning $\mathbb{N}$:**
   Since $\mathbb{N}$ is partitioned into $n$ disjoint sets, and one of these sets is a singleton, the remaining $n-1$ sets must cover the rest of $\mathbb{N}$.

5. **Finding the Greatest Singleton:**
   To find the greatest positive integer that may belong to $A_j$, we need to consider the structure of the other sets. If $A_i$ contains more than one element, it must contain all multiples of the gcd of any two of its elements. Therefore, the largest possible singleton $k$ must be such that it cannot be expressed as a sum of any two elements from the other sets.

6. **Using the Frobenius Coin Problem:**
   The Frobenius coin problem states that for two coprime integers $p$ and $q$, the largest integer that cannot be expressed as $ap + bq$ for non-negative integers $a$ and $b$ is $pq - p - q$. However, since we are dealing with positive integers, we need to consider $pq - p - q + 1$.

7. **Conclusion:**
   Given that $|A_j| = 1$, the largest integer that cannot be expressed as a sum of any two elements from the other sets must be the largest possible singleton. This is given by the Frobenius number for the remaining sets.

The final answer is $\boxed{n-1}$