Amazing Circle.

If $f(x,0) = 0$ has both roots equal to $x = 2$ , then $x^2 + 2ax + c = (x-2)^2 = x^2 - 4x + 4 = 0 \Rightarrow a = -2, c = 4.$ If $f(0,y) = 0$ has only integral roots, then $y^2 + 2by + 4 = 0,$ or:

$y = \frac{-2b \pm \sqrt{4b^2 - 16}}{2} = -b \pm \sqrt{b^2 - 4}$ (i).

We require the discriminant in (i) to be a perfect square if $y \in \mathbb{Z}.$ Thus, $b^2 - 4 = N^2 \Rightarrow b^2 - N^2 = (b+N)(b-N) = 4.$ The positive divisors of 4 are just 1, 2, or 4, which testing in each factor gives:

$b+N = 2, b-N = 2 \Rightarrow b = 2$ (ii)

$b+N = 4, b-N = 1 \Rightarrow b = \frac{5}{2}$ (iii).

Substituting (ii) and (iii) back into (i) yields:

$b = 2 \Rightarrow y = -2 \pm \sqrt{4 - 4} = -2$

$b = \frac{5}{2} \Rightarrow y = -\frac{5}{2} \pm \sqrt{\frac{25}{4} - 4} = -\frac{5}{2} \pm \frac{3}{2} = -1, -4$ .

So the sum of all positive $b$ equals $2 + \frac{5}{2} = \boxed{4.5}.$