1. Understanding the Problem:
We need to determine all finite nonempty sets $S$ of positive integers such that for any $i, j \in S$, the expression $\frac{i+j}{(i,j)}$ is also an element of $S$. Here, $(i,j)$ denotes the greatest common divisor (gcd) of $i$ and $j$.

2. Initial Observations:
- If $i = j$, then $\frac{i+i}{(i,i)} = \frac{2i}{i} = 2$, which implies that $2 \in S$ if $S$ contains any repeated element.
- If $i \neq j$, then $\frac{i+j}{(i,j)}$ must be an integer and an element of $S$.

3. Lemma 1:
- Define $f(i,j) = \frac{i+j}{(i,j)}$.
- We need to check if $f(i,j)$ is involutive, i.e., if $f(i,j) = i$ or $f(i,j) = j$.

4. Case Analysis:
- **Case 1: $\gcd(i,j) = 1$:**
- Here, $f(i,j) = i + j$.
- Since $i$ and $j$ are distinct, $i + j$ is greater than both $i$ and $j$. This implies that $S$ must contain larger elements, leading to an infinite set, which contradicts the finiteness of $S$.

- **Case 2: $\gcd(i,j) = k > 1$:**
- Let $i = k \cdot i'$ and $j = k \cdot j'$ where $\gcd(i', j') = 1$.
- Then, $f(i,j) = \frac{k(i' + j')}{k} = i' + j'$.
- Since $i'$ and $j'$ are coprime, $i' + j'$ must be an element of $S$.

5. Finite Set Construction:
- Consider $S = \{ n \}$ for any positive integer $n$. This trivially satisfies the condition since there are no distinct pairs $i, j$ to consider.
- Consider $S = \{ n, n(n-1) \}$ for $n > 2$:
- For $i = n$ and $j = n(n-1)$, we have $\gcd(n, n(n-1)) = n$.
- Then, $f(n, n(n-1)) = \frac{n + n(n-1)}{n} = \frac{n^2}{n} = n$.
- This implies $n \in S$, which is already true.

6. Verification:
- For $S = \{ n \}$, the condition is trivially satisfied.
- For $S = \{ n, n(n-1) \}$, the condition is satisfied as shown above.

7. Conclusion:
- The only finite nonempty sets $S$ that satisfy the given condition are of the form $S = \{ n \}$ and $S = \{ n, n(n-1) \}$ for $n > 2$.

The final answer is $\boxed{ S = \{ n \} }$ or $S = \{ n, n(n-1) \}$ for $n > 2$.