If we look at $\mathbb{N}\subset \mathbb{R}$ every point is not a limit point as the intersection of neighborhood of the point $p\in \mathbb{N}$ and $\mathbb{N}$ is empty.
But we we look at $\mathbb{N}$ not as the subset of $\mathbb{R}$ but as the whole space, can we find limit points in $\mathbb{N}$?
In order to answer a question about limit points in a set, you first need to define the "topology" on the set. In your case, this likely amounts a metric on the set $\mathbb N$.
The standard metric on $\mathbb N$ is the same metric you get when you consider $\mathbb N$ as a subspace o $\mathbb R$ - namely, $d(m,n)=|m-n|$. Under this metric, by the same reasoning as above, given $m$, there is no $n\neq m$ such that $d(m,n)<\frac{1}{2}$, so $m$ is not a limit point of $\mathbb N$.
There are lots of other metrics on $\mathbb N$. Given any one-to-one map $g:\mathbb N\to\mathbb R$, you can define $d(m,n)=|g(m)-g(n)|$, and that is a metric. In particular, there is a one-to-one and onto map $g:\mathbb N\to \mathbb Q$, and, with the metric from this map, every point is a limit point.
The most interesting non-standard metrics on the naturals, though, are probably the $p$-adic metrics. Indeed, there is a sense in which the standard metric is the $0$-adic metric.
There are topologies on $\mathbb N$ that do not come from metrics, but since you only listed "metric spaces" in your tags, I won't cover that.
No, in any compatible metric. All $n \in \mathbb{N}$ will be isolated points. An isolated point isn't a limit point of anything.
