$N?$

It is easy to see that when $a=b=c=1$ , then $N=3$ and it satisfies the equation.

Now, we prove that this is the only possibility.

First of all, we use the identity

$(ab+bc+ca)^{2} \geq 3abc(a+b+c)$

[This can be proved by using the identity $x^{2}+y^{2}+z^{2} \geq xy+yz+zx$ , and then adding $2(xy+yz+zx)$ to both sides to get

$(x+y+z)^{2} \geq 3(xy+yz+zx)$ and then substituting $x=ab, y=bc$ and $z=ca$ ]

$\implies ab+bc+ca \geq 3abc$ [because $ab+bc+ca=a+b+c>0$ ]

$\implies \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq3$

But since $a \geq 1$ , thus $\dfrac{1}{a} \leq 1$ and similarly two more results. Adding them gives

$3 \geq \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$
And hence by these two results, we conclude that equality holds and thus $a=b=c=1$ is the only possibility which gives $N=3$ .

Hence the sum of all possible values is $\boxed {3}$