Suppose you have a topology on set of integers (say the set of all positive integers) that are neither discrete nor indiscrete. And suppose we also do not assume it to be Hausdorff.
Is it possible to construct the homotopies, and define fundamental group or groupoid on the such topology on integers? I cannot imagine constructing the homotopies of paths given that the open sets and topologies are based on the integers...
On
You can always define them.
In fact, there are even finite sets with weird topologies that have nontrivial homotopy groups; so where the notions are in fact interesting ! (and of course well-defined)
If you look at, e.g. the first paragraph of Chapter 3 here, you will see that for any finite simplicial complex, there is a finite topological space which is weakly equivalent, so in particular it has the same homotopy groups.
Let's take an example, as she describes the topology in question. Let's say we want to model $S^1$, the circle. Let's take a simplicial complex that models it, e.g. the usual cell-structure with three points and three edges between them. We thus have $6$ points, $v_0,v_1,v_2,e_0,e_1,e_2$, with $v_0,v_1\leq e_2; v_0,v_2\leq e_1; v_1,v_2\leq e_0$ and nothing else.
The order topology has basic open sets of the form $U_y = \{x\mid x\leq y\}$ for all $y$. Here, we therefore have a bunch of open sets $\{e_i,v_j,v_k\}$for $\{i,j,k\}=\{0,1,2\}$, and $\{v_i\}$ for any $i$. Let's call our space $X$
First of all, this is path-connected. We'll prove this by showing that $e_0$ and $v_1$ are connected by a path, which, by symmetry and composition of paths, will be enough .
Let $\gamma :[0,1]\to X$ be defined by $\gamma(t) = v_1$ for $t<1$, and $\gamma(1) = e_0$. It's easy to check that this is continuous.
How about the fundamental group ? Well, I won't prove that the loop I give is nontrivial, but let me just explicitly describe a nontrivial loop is : go from $v_1$ to $e_0$ as before, then from $e_0$ to $v_2$ in the same way, then from $v_2$ to $e_1$, then from $e_1$ to $v_0$, $v_0$ to $e_2$, and finally $e_2$ to $v_1$.
Note that $X$ is path-connected, locally path-connected, even locally contractible, so you can apply most of covering space theory to $X$. In particular, you can prove that the loop I described is nontrivial (and actually is torsion-free) by exhibiting a covering space where it lifts as a path that isn't a loop (and work a bit more for torsion-free, just like for $S^1$). To exhibit covering spaces like that, just use the intuition that this space is "just $S^1$".
I'm not an expert on algebraic topology, but I'll point out that the usual definition of the fundamental group is extremely uninteresting if the space is countable and the topology is at least $T_1$.
Let $X$ be your space - for this discussion it doesn't matter that the underlying set is $\mathbb{Z}$, but it does matter that $X$ is countable. Pick a basepoint $x_0 \in X$. A loop based at $x_0$ is usually defined as a continuous map $f : [0, 1] \to X$ such that $f(0) = f(1) = x_0$. Since $[0, 1]$ is path-connected, and the continuous image of a path-connected space is also path-connected, the image of $f$ must be path-connected. A path-connected $T_1$ space with at least two points must be uncountable, so the image of $f$ must be a single point. Thus the only loop based at $x_0$ is the constant loop, and the fundamental group is trivial.
I don't have an immediate answer for what happens if the topology is not even $T_1$.