Find all integer triples $(a,b,c)$ with $a>0>b>c$ whose sum equal $0$ such that the number
$$N=2017-a^3b-b^3c-c^3a$$ is a perfect square of an integer.
To find all integer triples \((a, b, c)\) with \(a > 0 > b > c\) whose sum equals \(0\) such that the number 
\[ N = 2017 - a^3b - b^3c - c^3a \]
is a perfect square of an integer, we proceed as follows:

1. **Substitute \(d = -b\) and \(e = -c\):**
   \[
   a + b + c = 0 \implies a - d - e = 0 \implies a = d + e
   \]
   Thus, we have \(a = d + e\).

2. **Rewrite \(N\) in terms of \(d\) and \(e\):**
   \[
   N = 2017 - a^3b - b^3c - c^3a
   \]
   Substituting \(a = d + e\), \(b = -d\), and \(c = -e\), we get:
   \[
   N = 2017 - (d+e)^3(-d) - (-d)^3(-e) - (-e)^3(d+e)
   \]
   Simplifying each term:
   \[
   (d+e)^3(-d) = -(d^3 + 3d^2e + 3de^2 + e^3)d = -d^4 - 3d^3e - 3d^2e^2 - de^3
   \]
   \[
   (-d)^3(-e) = d^3e
   \]
   \[
   (-e)^3(d+e) = -e^3(d+e) = -e^3d - e^4
   \]
   Combining these, we get:
   \[
   N = 2017 - (-d^4 - 3d^3e - 3d^2e^2 - de^3 + d^3e - e^3d - e^4)
   \]
   \[
   N = 2017 + d^4 + 3d^3e + 3d^2e^2 + de^3 - d^3e + e^3d + e^4
   \]
   Simplifying further:
   \[
   N = 2017 + d^4 + 2d^3e + 3d^2e^2 + 2de^3 + e^4
   \]

3. **Express \(N\) as a perfect square:**
   \[
   N = 2017 + (d^2 + e^2 + de)^2
   \]
   Let \(s = d^2 + e^2 + de\). Then:
   \[
   N = 2017 + s^2
   \]
   For \(N\) to be a perfect square, we need:
   \[
   2017 + s^2 = x^2 \implies x^2 - s^2 = 2017 \implies (x+s)(x-s) = 2017
   \]
   Since 2017 is a prime number, the only factor pairs are \((1, 2017)\) and \((-1, -2017)\). Thus, we have:
   \[
   x + s = 2017 \quad \text{and} \quad x - s = 1 \quad \text{or} \quad x + s = -1 \quad \text{and} \quad x - s = -2017
   \]
   Solving for \(x\) and \(s\):
   \[
   x + s = 2017 \quad \text{and} \quad x - s = 1 \implies 2x = 2018 \implies x = 1009, \quad 2s = 2016 \implies s = 1008
   \]
   \[
   x + s = -1 \quad \text{and} \quad x - s = -2017 \implies 2x = -2018 \implies x = -1009, \quad 2s = -2016 \implies s = -1008
   \]
   Since \(s = d^2 + e^2 + de > 0\), we discard the negative solution. Thus, \(s = 1008\).

4. **Solve \(d^2 + e^2 + de = 1008\):**
   Let \(d = 2^m d_0\) and \(e = 2^n e_0\) with \(d_0\) and \(e_0\) being odd. Since \(1008 = 2^4 \cdot 63\), we need \( \min(2m, 2n) = 4\).

   Assume \(m = 2\):
   \[
   d = 4d_0 \quad \text{and} \quad e = 2^n e_0
   \]
   Dividing both sides by \(16\):
   \[
   d_0^2 + t^2 + t d_0 = 63 \quad \text{where} \quad t = \frac{e}{4}
   \]
   Since \(d_0\) is odd, we test \(d_0 = 1, 3, 5, 7\):
   - For \(d_0 = 1\):
     \[
     1 + t^2 + t = 63 \implies t^2 + t - 62 = 0 \implies t = \frac{-1 \pm \sqrt{1 + 248}}{2} = \frac{-1 \pm 15}{2}
     \]
     \[
     t = 7 \quad \text{or} \quad t = -8
     \]
     Thus, \(e = 4t = 28\) or \(-32\), but \(e\) must be positive, so \(e = 28\).

   - For \(d_0 = 3\):
     \[
     9 + t^2 + 3t = 63 \implies t^2 + 3t - 54 = 0 \implies t = \frac{-3 \pm \sqrt{9 + 216}}{2} = \frac{-3 \pm 15}{2}
     \]
     \[
     t = 6 \quad \text{or} \quad t = -9
     \]
     Thus, \(e = 4t = 24\) or \(-36\), but \(e\) must be positive, so \(e = 24\).

   - For \(d_0 = 5\) and \(d_0 = 7\), no integer solutions for \(t\).

   Therefore, the only valid solution is \(d_0 = 3\), \(t = 6\), \(e = 24\), and \(d = 12\).

5. **Find \(a, b, c\):**
   \[
   a = d + e = 12 + 24 = 36
   \]
   \[
   b = -d = -12
   \]
   \[
   c = -e = -24
   \]

The final answer is \(\boxed{(36, -12, -24)}\).