Consider an absolute value equation for $a, b\in \mathbb{R}$ and $c\in \mathbb{N}$,
$$|x+a|+|x+b| = |c|$$
What can be said about the number of solutions to this equation?
If $x+a<0$ and $x+b>0$, the equation holds for every $x\in \mathbb{R}$ if and only if
$$b-a = c\tag{1}$$
If $x+a>0$ and $x+b>0$, the equation has a constant solution
$$x = C(a, b, c)\tag{2}$$
If $x+a<0$ and $x+b<0$, the equation again has a constant solution
$$x = C(a, b, c)\tag{3}$$
Lastly, if $x+a>0$ and $x+b<0$, the equation holds for every $x\in \mathbb{R}$ if and only if
$$a-b = c\tag{4}$$
But there exist $a,b\in \mathbb{R}$ such that $a-b, b-a\notin\mathbb{N}$, and so $(1)$ and $(4)$ won't hold for arbitrary $a, b, c$.
$a$ and $b$ are being treated as parameters here, so the number of solutions can vary in the different cases that arise according to the values of $a$ and $b$. In other words, you're asked to find the number of solutions "as a function of" $a$ and $b$, rather than specifically finding the cases in which the number of solutions is the same for all arbitrary $a$ and $b$.
If $x+a \leq 0$ and $x+b \geq 0$, i.e. $-b \leq x \leq -a$, then the equation becomes $b-a = c$. That would mean $b = a+c > a$ (assuming you say $0 \not\in \mathbb{N}$), so $-b < -a$, meaning that there are infinitely many $x$-values in the required value. Hence this case gives infinitely many solutions if $b-a = c$, and otherwise doesn't give any solutions.
Analogously, if $x+a \geq 0$ and $x+b \leq 0$, i.e. $-a \leq x \leq -b$, then the equation becomes $a-b = c$. This similarly would imply $a > b$, so $-a < -b$, so there are infinitely many solutions in this case if $a-b = c$, and otherwise this case gives no solutions.
Putting these two cases together, we've shown so far that if $c = b-a$ or $c = a-b$, i.e. if $c = \left\lvert a-b \right\rvert$, then there are infinitely many solutions.
If $x+a > 0$ and $x+b > 0$, we must have $x > -a$ and $x > -b$, and the solution would be $x = \frac{c-a-b}{2}$. Hence this case gives one solution if $$\begin{align*}&\frac{c-a-b}{2} > -a \iff c-a-b > -2a \iff c > b-a, \text{ and} \\ &\frac{c-a-b}{2} > -b \iff c-a-b > -2b \iff c > a-b,\end{align*}$$ i.e. if $c > \left\lvert a-b\right\rvert$.
By the same argument, from the $x+a < 0$ and $x+b < 0$ case, we also get one solution, namely $x = \frac{-c-a-b}{2}$, if $c > \left\lvert a-b\right\rvert$. As $c \neq 0$, this value of $x$ is different from that found above, so we conclude that: