For which $n \ge 2$ can one find a sequence of distinct positive integers $a_1, a_2, \ldots , a_n$ so that the sum
$$\frac{a_1}{a_2} + \frac{a_2}{a_3} + \ldots +\frac{a_n}{a_1}$$
 is an integer?

[i](3 points)[/i]
1. **Case \( n = 2 \):**
   - Let \( a_1 \) and \( a_2 \) be distinct positive integers.
   - Let \( d = \gcd(a_1, a_2) \), then \( a_1 = dn \) and \( a_2 = dm \) where \( \gcd(n, m) = 1 \).
   - Consider the sum:
     \[
     \frac{a_1}{a_2} + \frac{a_2}{a_1} = \frac{dn}{dm} + \frac{dm}{dn} = \frac{n}{m} + \frac{m}{n}
     \]
   - Simplify the expression:
     \[
     \frac{n}{m} + \frac{m}{n} = \frac{n^2 + m^2}{mn}
     \]
   - For this to be an integer, \( mn \) must divide \( n^2 + m^2 \). Since \( \gcd(n, m) = 1 \), \( mn \) divides \( (n+m)^2 - 2nm \):
     \[
     \frac{n^2 + m^2}{mn} = \frac{(n+m)^2 - 2nm}{mn} = \frac{(n+m)^2}{mn} - 2
     \]
   - Since \( n \) and \( m \) are coprime, \( (n+m, n) = (n+m, m) = 1 \). Therefore, \( mn \) divides \( (n+m)^2 \) only if \( n = m = 1 \), which contradicts the requirement that \( a_1 \) and \( a_2 \) are distinct. Hence, no such sequence exists for \( n = 2 \).

2. **Case \( n = 3 \):**
   - Consider the sequence \( (1, 2, 4) \).
   - Calculate the sum:
     \[
     \frac{1}{2} + \frac{2}{4} + \frac{4}{1} = \frac{1}{2} + \frac{1}{2} + 4 = 5
     \]
   - The sum is an integer, so such a sequence exists for \( n = 3 \).

3. **Case \( n > 3 \):**
   - Consider the sequence \( a_1 = a_2 = \ldots = a_{n-2} = 1 \), \( a_{n-1} = 2 \), \( a_n = 4 \).
   - Calculate the sum:
     \[
     \frac{a_1}{a_2} + \frac{a_2}{a_3} + \ldots + \frac{a_{n-2}}{a_{n-1}} + \frac{a_{n-1}}{a_n} + \frac{a_n}{a_1}
     \]
   - Since \( a_1 = a_2 = \ldots = a_{n-2} = 1 \), the sum simplifies to:
     \[
     \frac{1}{1} + \frac{1}{1} + \ldots + \frac{1}{1} + \frac{1}{2} + \frac{2}{4} + \frac{4}{1}
     \]
   - There are \( n-3 \) terms of \( \frac{1}{1} \), so the sum is:
     \[
     (n-3) \cdot 1 + \frac{1}{2} + \frac{1}{2} + 4 = (n-3) + 1 + 4 = n + 2
     \]
   - The sum is an integer, so such a sequence exists for \( n > 3 \).

The final answer is \( \boxed{ n = 3 } \) or \( n > 3 \).