Actually, it is a partial order. Given natural numbers $a,b$, we can conclude from $a\vert b\wedge b\vert a$ that $a=b.$ Your counterexample doesn't work, because, while everything divides $0$, $0$ only divides itself.
A maximum element would be some $(a,b)\in T$ such that for all $(c,d)\in T$, $c\leq a\wedge d\vert b$. This would require that $a$ be the greatest element of $\Bbb Z^-$ and that $b$ be a natural number that all natural numbers divide. Can you think of what that would be? Note that if there is a maximum element, then it is the only maximal element.
A minimum element would be some $(a,b)\in T$ such that for all $(c,d)\in T$, $c\geq a\wedge b\vert d$. Then $a$ would have to be the least element of $\Bbb Z^-$, which is not possible.
Given any $(a,b)$, we have that $(a-1,b)\odot(a,b)$, so there is always a different element preceding it. Thus, there are no minimal elements, either.
To determine whether or not $\odot$ totally orders $T$, we must determine whether for distinct $(a,b),(c,d)\in T$ we need have $(a,b)\odot(c,d)$ or $(c,d)\odot(a,b)$. It shouldn't be difficult to see that this need not hold (I leave it to you to find an example), so that $\odot$ does not totally order $T$, after all.
Edit: Well done in showing that it isn't a total order, and in finding the maximum element. It is in fact maximum (not just maximal) since for each $(c,d)\in T$ we have $c\leq-1$ and $d\vert 0$. Partial orders may or may not have a maximum or a minimum element, and may or may not have maximal or minimal elements. For an example of a partial order with a maximum element, take any set $A$, any $b\notin A$ and define a partial order $\precsim$ on $A\cup\{b\}$ by $c\precsim c$ for all $c$, and $a\precsim b$ for all $a\in A$. Then $b$ is the maximum element, and each $a\in A$ is incomparable to the others, so minimal, but not minimum if $A$ has multiple elements.
To elaborate on the antisymmetry of your example, let's suppose that $m,n\in\Bbb N$ such that $m\vert n\wedge n\vert m$. Hence, $$m=nv\wedge n=mw$$ for some $v,w\in\Bbb N$. If $m=0$, then so is $n$. If $m\neq 0$, then since $m=nv=mvw$, then $vw=1$, which is only possible for $v=w=1$ since $v,w\in\Bbb N$. Thus, again, $m=n$. Does that clear things up?
Best Answer
If we supposed that Love was a partial order then we immediately run into issues with anti-symmetry. To be precise; if $x$ loves $y$ and $y$ loves $x$ we have the equations $x \leq y$ and $y \leq x$, thus by the anti-symmetry axiom we have $x=y$ and so $x$ and $y$ are the same person. Assuming that you don't take the narcissistic view that people only love themselves I would say this does not hold always. A total order would also run into the same problem.
As for it being an equivalence relation I would say no as it fails on all 3 criteria:
If I had to categorise I would say your best bet is to think of Love being a directed graph with People as the vertices and an edge from $a$ to $b$ representing $a$ loves $b$.