Let $a,b,c$ be three distinct positive integers such that the sum of any two of them is a perfect square and having minimal sum $a + b + c$. Find this sum.
1. We start with the given conditions:
   \[
   a + b = p^2, \quad b + c = q^2, \quad a + c = r^2
   \]
   where \( p, q, r \) are integers.

2. Adding all three equations, we get:
   \[
   (a + b) + (b + c) + (a + c) = p^2 + q^2 + r^2
   \]
   Simplifying, we obtain:
   \[
   2a + 2b + 2c = p^2 + q^2 + r^2
   \]
   Dividing both sides by 2, we have:
   \[
   a + b + c = \frac{p^2 + q^2 + r^2}{2}
   \]

3. For \( a, b, c \) to be integers, \(\frac{p^2 + q^2 + r^2}{2}\) must also be an integer. This implies that \( p^2 + q^2 + r^2 \) must be even. Therefore, \( p, q, r \) must be such that the sum of their squares is even. This means that either all three are even, or one is even and two are odd.

4. To minimize \( a + b + c \), we should choose the smallest possible values for \( p, q, r \) such that \( p^2, q^2, r^2 \) can form a triangle (triangle inequality must hold).

5. Let's try small values for \( p, q, r \):
   - \( p = 5, q = 6, r = 7 \)
   - Check the triangle inequality:
     \[
     5^2 + 6^2 > 7^2 \quad \text{(25 + 36 > 49, true)}
     \]
     \[
     5^2 + 7^2 > 6^2 \quad \text{(25 + 49 > 36, true)}
     \]
     \[
     6^2 + 7^2 > 5^2 \quad \text{(36 + 49 > 25, true)}
     \]
   - All conditions are satisfied.

6. Calculate \( a, b, c \):
   \[
   a + b = 5^2 = 25
   \]
   \[
   b + c = 6^2 = 36
   \]
   \[
   a + c = 7^2 = 49
   \]

7. Solve for \( a, b, c \):
   \[
   a + b = 25 \quad \text{(1)}
   \]
   \[
   b + c = 36 \quad \text{(2)}
   \]
   \[
   a + c = 49 \quad \text{(3)}
   \]

   Adding equations (1) and (2):
   \[
   a + b + b + c = 25 + 36
   \]
   \[
   a + 2b + c = 61 \quad \text{(4)}
   \]

   Subtracting equation (3) from equation (4):
   \[
   a + 2b + c - (a + c) = 61 - 49
   \]
   \[
   2b = 12
   \]
   \[
   b = 6
   \]

   Substitute \( b = 6 \) into equation (1):
   \[
   a + 6 = 25
   \]
   \[
   a = 19
   \]

   Substitute \( b = 6 \) into equation (2):
   \[
   6 + c = 36
   \]
   \[
   c = 30
   \]

8. Therefore, the minimal sum \( a + b + c \) is:
   \[
   a + b + c = 19 + 6 + 30 = 55
   \]

The final answer is \(\boxed{55}\)