The positive integers $a, b, c$ are such that
$$gcd \,\,\, (a, b, c) = 1,$$
$$gcd \,\,\,(a, b + c) > 1,$$
$$gcd \,\,\,(b, c + a) > 1,$$
$$gcd \,\,\,(c, a + b) > 1.$$
Determine the smallest possible value of $a + b + c$.

Clarification: gcd stands for greatest common divisor.
1. Given the conditions:
   \[
   \gcd(a, b, c) = 1,
   \]
   \[
   \gcd(a, b + c) > 1,
   \]
   \[
   \gcd(b, c + a) > 1,
   \]
   \[
   \gcd(c, a + b) > 1,
   \]
   we need to determine the smallest possible value of \(a + b + c\).

2. Let's denote \(k = a + b + c\). From the conditions, we have:
   \[
   \gcd(a, k - a) > 1,
   \]
   \[
   \gcd(b, k - b) > 1,
   \]
   \[
   \gcd(c, k - c) > 1.
   \]
   This implies that \(k - a\), \(k - b\), and \(k - c\) must share a common factor with \(a\), \(b\), and \(c\) respectively, which is greater than 1.

3. Since \(\gcd(a, b, c) = 1\), \(a\), \(b\), and \(c\) must be pairwise coprime. This means that \(k\) must be divisible by at least three distinct prime factors to satisfy the conditions.

4. Let's consider the smallest possible value of \(k\) that has three distinct prime factors. The smallest such number is \(2 \times 3 \times 5 = 30\).

5. We need to find \(a\), \(b\), and \(c\) such that:
   \[
   a + b + c = 30,
   \]
   and the conditions \(\gcd(a, b + c) > 1\), \(\gcd(b, c + a) > 1\), and \(\gcd(c, a + b) > 1\) are satisfied.

6. Let's try \(a = 2\), \(b = 3\), and \(c = 25\):
   \[
   a + b + c = 2 + 3 + 25 = 30.
   \]
   Check the conditions:
   \[
   \gcd(2, 3 + 25) = \gcd(2, 28) = 2 > 1,
   \]
   \[
   \gcd(3, 25 + 2) = \gcd(3, 27) = 3 > 1,
   \]
   \[
   \gcd(25, 2 + 3) = \gcd(25, 5) = 5 > 1.
   \]
   All conditions are satisfied.

The final answer is \(\boxed{30}\).