Find all quadruples $(a, b, c, d)$ of positive integers satisfying $\gcd(a, b, c, d) = 1$ and
\[ a | b + c, ~ b | c + d, ~ c | d + a, ~ d | a + b. \]

[i]Vítězslav Kala (Czech Republic)[/i]
To find all quadruples \((a, b, c, d)\) of positive integers satisfying \(\gcd(a, b, c, d) = 1\) and the conditions:
\[ a \mid (b + c), \quad b \mid (c + d), \quad c \mid (d + a), \quad d \mid (a + b), \]
we will proceed by considering different cases based on the relative magnitudes of \(a, b, c,\) and \(d\).

1. **Assume \(a = \max(a, b, c, d)\):**
   - Since \(a \mid (b + c)\), we have \(\frac{b + c}{a} \in \{1, 2\}\).

2. **Case 1: \(b + c = 2a\):**
   - If \(b + c = 2a\), then \(a = b = c\). 
   - Since \(a \mid d\), and \(\gcd(a, b, c, d) = 1\), the only possibility is \(a = b = c = d = 1\).
   - Thus, \(\boxed{(1, 1, 1, 1)}\) is a solution.

3. **Case 2: \(b + c = a\):**
   - **Subcase 2.1: \(b = \max(b, c, d)\):**
     - Since \(b \mid (c + d)\), we have \(\frac{c + d}{b} \in \{1, 2\}\).
     - If \(c + d = 2b\), then \(a = 2b\) and \(b = c = d = 1\).
     - Thus, \(\boxed{(2, 1, 1, 1)}\) is a solution.
     - If \(c + d = b\), then \(c \mid 2d\) and \(d \mid 3c\) with \(\gcd(c, d) = 1\).
     - Hence, \(c \mid 2\) and \(d \mid 3\), leading to the solutions:
       \[
       \boxed{(3, 2, 1, 1)}, \quad \boxed{(5, 3, 2, 1)}, \quad \boxed{(5, 4, 1, 3)}, \quad \boxed{(7, 5, 2, 3)}
       \]

   - **Subcase 2.2: \(c = \max(b, c, d)\) and \(c > b\):**
     - Since \(c \mid (d + a)\), we have \(\frac{d + a}{c} \in \{1, 2\}\).
     - If \(d + a = 2c\), then \(a = 2c\) and \(b = d = 1\).
     - Thus, \(\boxed{(3, 1, 2, 1)}\) is a solution.
     - If \(d + a = c\), then \(b \mid 2d\) and \(d \mid 3b\) with \(\gcd(b, d) = 1\).
     - Hence, \(b \mid 2\) and \(d \mid 3\), leading to the solutions:
       \[
       \boxed{(5, 1, 4, 3)}, \quad \boxed{(5, 2, 3, 1)}, \quad \boxed{(7, 2, 5, 3)}
       \]

   - **Subcase 2.3: \(d = \max(b, c, d)\):**
     - Since \(d \mid (a + b)\), we have \(\frac{a + b}{d} \in \{1, 2\}\).
     - If \(a + b = 2d\), then \(a = 2d\) and \(b = c = 1\).
     - Thus, \(\boxed{(7, 3, 4, 5)}\) is a solution.

4. **Permutations:**
   - All the solutions found can be permuted to generate additional solutions.