Find all subsets $A$ of $\left\{ 1, 2, 3, 4, \ldots \right\}$, with $|A| \geq 2$, such that for all $x,y \in A, \, x \neq y$, we have that $\frac{x+y}{\gcd (x,y)}\in A$.

[i]Dan Schwarz[/i]
1. **Initial Assumptions and Definitions:**
   - Let \( A \) be a subset of \( \{1, 2, 3, 4, \ldots\} \) with \( |A| \geq 2 \).
   - For all \( x, y \in A \) with \( x \neq y \), we have \( \frac{x+y}{\gcd(x,y)} \in A \).

2. **Case 1: \( |A| = 2 \):**
   - Let \( A = \{a, b\} \) with \( a < b \).
   - Assume \( \gcd(a, b) = d > 1 \).
   - Let \( a = dx \) and \( b = dy \).
   - From the hypothesis, \( \frac{a+b}{\gcd(a,b)} = \frac{dx + dy}{d} = x + y \in A \).
   - Since \( x + y \leq \frac{a+b}{2} < b \), it must be that \( x + y = a \), i.e., \( x + y = dx \).
   - This implies \( y = x(d-1) \), meaning \( x \mid y \).
   - Given \( \gcd(x, y) = 1 \), it follows that \( x = 1 \) and \( y = d-1 \).
   - Thus, \( a = d \) and \( b = d(d-1) \), leading to the solution \( A = \{d, d(d-1)\} \) for \( d \geq 3 \).

3. **Case 2: \( |A| > 2 \):**
   - Let \( a < b < c \) be the smallest elements of \( A \).
   - Assume \( \gcd(a, c) > 1 \) and \( \gcd(b, c) > 1 \).
   - Let \( e = \gcd(a, c) \) and \( a = eu \), \( c = ev \).
   - From the hypothesis, \( \frac{a+c}{\gcd(a,c)} = \frac{eu + ev}{e} = u + v \in A \).
   - Since \( u + v < c \), it must be one of the smaller elements, i.e., \( u + v = a \) or \( u + v = b \).
   - If \( u + v = a \), then \( v = a - u \).
   - If \( u + v = b \), then \( v = b - u \).
   - Since \( \gcd(a, c) = e \), we have \( u \mid v \) and \( \gcd(u, v) = 1 \).
   - This implies \( u = 1 \) and \( v = e(a-1) - 1 \).
   - Thus, \( \gcd(b, c) = \gcd(a, c) = d \), and \( b = \frac{a+c}{\gcd(a,c)} < \frac{b+c}{\gcd(b,c)} < c \), leading to a contradiction.
   - Therefore, if \( |A| \geq 3 \), there must exist two coprime elements in \( A \).

4. **Conclusion:**
   - The only subsets \( A \) of \( \{1, 2, 3, 4, \ldots\} \) with \( |A| \geq 2 \) that satisfy the given condition are of the form \( \{d, d(d-1)\} \) for \( d \geq 3 \).

The final answer is \( \boxed{ \{d, d(d-1)\} } \) for \( d \geq 3 \).