Find all pairs $(a, b)$ of positive integers for which $a + b = \phi (a) +  \phi (b) + gcd (a, b)$.

Here $ \phi (n)$ is the number of numbers $k$ from $\{1, 2,. . . , n\}$ with $gcd (n, k) = 1$.
To solve the problem, we need to find all pairs \((a, b)\) of positive integers such that the equation \(a + b = \phi(a) + \phi(b) + \gcd(a, b)\) holds. Here, \(\phi(n)\) is Euler's totient function, which counts the number of integers up to \(n\) that are coprime with \(n\).

1. **Express \(a\) and \(b\) in terms of their greatest common divisor:**
   Let \(d = \gcd(a, b)\). Then we can write \(a = dx\) and \(b = dy\) where \(\gcd(x, y) = 1\). Substituting these into the given equation, we get:
   \[
   dx + dy = \phi(dx) + \phi(dy) + d
   \]

2. **Use properties of the totient function:**
   Recall that \(\phi(dx) = \phi(d) \cdot \phi(x)\) when \(\gcd(d, x) = 1\). Since \(\gcd(x, y) = 1\), we have:
   \[
   \phi(dx) = \phi(d) \cdot \phi(x) \quad \text{and} \quad \phi(dy) = \phi(d) \cdot \phi(y)
   \]
   Substituting these into the equation, we get:
   \[
   dx + dy = \phi(d) \cdot \phi(x) + \phi(d) \cdot \phi(y) + d
   \]
   Factoring out \(\phi(d)\), we obtain:
   \[
   d(x + y) = \phi(d) (\phi(x) + \phi(y)) + d
   \]
   Simplifying, we get:
   \[
   d(x + y - 1) = \phi(d) (\phi(x) + \phi(y))
   \]

3. **Analyze the equation:**
   We need to find when the equality \(d(x + y - 1) = \phi(d) (\phi(x) + \phi(y))\) holds. Consider the following cases:

   - **Case 1: \(d = 1\):**
     If \(d = 1\), the equation simplifies to:
     \[
     x + y - 1 = \phi(x) + \phi(y)
     \]
     Assume \(x\) and \(y\) are composite. For composite numbers \(x\) and \(y\), \(\phi(x) \leq x - 1\) and \(\phi(y) \leq y - 1\). This implies:
     \[
     x + y - 1 \leq (x - 1) + (y - 1) = x + y - 2
     \]
     This is a contradiction unless \(x\) or \(y\) is 1. Therefore, \(x\) and \(y\) must be either 1 or prime. This gives us the solutions \((1, p)\) and \((p, 1)\) where \(p\) is a prime number.

   - **Case 2: \(d \neq 1\):**
     If \(d \neq 1\), we need to find when \(d(x + y - 1) = \phi(d) (\phi(x) + \phi(y))\) holds. Assume \(x\) and \(y\) are composite. Then \(\phi(x) \leq x - 1\) and \(\phi(y) \leq y - 1\). This implies:
     \[
     x + y - 1 \leq \frac{x + y}{2}
     \]
     This is a contradiction unless \(x = y = 1\). Therefore, \(x\) and \(y\) must be either 1 or prime. This gives us the solutions \((2^t, 2^t)\) where \(t \geq 1\).

4. **Conclusion:**
   The solutions are \((x, y) \in \{(2^t, 2^t); (1, p); (p, 1)\}\), where \(t \geq 1\) and \(p\) is a prime number.

The final answer is \( \boxed{ (x, y) \in \{(2^t, 2^t); (1, p); (p, 1)\} } \), where \(t \geq 1\) and \(p\) is a prime number.