Equivalence classes (wrt your equivalence relation) are subsets of elements of the original set with the following property: every element of a certain equivalence class must be equivalent (wrt the equivalence relation) to any other in it and inequivalent to any other out of it.
This implies that equivalence classes are disjoint subsets of the original set. (*)
You are asked to construct the maximal subsets of $\mathbb Z$, such that - picked one - every element in it has the same absolute value as any other element of the subset. It should be fairly easy to understand how many elements can belong to every equivalence class in this case (there is an important exception, though: $0$).
Notice that in general the cardinalities of the equivalence classes of a set wrt an equivalence relation are different!
(*) This second part of the definition was added after the useful comments of two users below.
(a)R in N will be reflexive iff (m,n) R (m,n).
Now, m+n=n+m$\implies$(m,n) R (m,n)$\implies$R is reflexive
(b)R in N will be symmetric if (m,n) R (l,k)$\implies$ (l,k) R (m,n).
Now, m+k=n+l$\implies$l+n=k+m$\implies$ (m,n) R (l,k)$\implies$ (l,k) R (m,n)$\implies$R is symmetric.
(c)R in N will be transitive if (m,n) R (l,k) and (l,k) R (p,q)$\implies$(m,n) R (p,q). Now,
(m,n) R (l,k) and (l,k) R (p,q) $\implies$ m+k=n+l and l+q=k+p$\implies$k=n+l-m and k=l+q-p
$\implies$n+l-m=l+q-p$\implies$n-m=q-p$\implies$n+p=q+m$\implies$m+q=n+p$\implies$(m,n) R (p,q). Thus, R is transitive.
Best Answer
Hint
$$ m^2 = n^2 \iff m^2 - n^2 = 0 \iff (m - n)(m + n) = 0 \iff m = n \text{ or } m = -n $$
So $m$ and $n$ are equivalent if and only if either $m = n$ or $m = -n$.