From page 5 of https://mast.queensu.ca/~kani/papers/numgenl.pdf :
Then $\psi$ is automatically an anti-isometry with respect to the $e_N$ -pairings on $E$ and on $E'$: $$e_N(\psi(x), \psi(y)) = e_N(x, y)^{-1}, \forall x, y \in J_E[N],$$ and this condition is equivalent to the assertion that the "graph subgroup scheme" $H_\psi = \text{Graph}(\psi) \leq (J_E \times J_{E'})[N]$ Is a (maximal) isotropic subgroup of $(J_E \times J_{E'})[N]$ (with respect to the $e_N$-pairing on $A = J_E \times J_{E'}$).
What does a maximal isotropic subgroup with respect to a Weil pairing mean? If possible I would like a reference to a definition.
Finding a maximal isotropic subspace : This question talks about isotropic subspaces, which might work if we consider the elliptic curve as a vector space.
If $A$ is a principally polarised abelian variety (so that we may equip $A[n]$ with the symplecitc Weil pairing $e_{A,n}$ taking values in $\mu_n$), then a subgroup $G \subset A[n]$ is said to be isotropic (wrt $e_{A,n}$) if $e_{A,n}$ is trivial on $G$ (i.e., everything in $G$ pairs to $1$). It is maximal isotropic if it is not properly contained in another isotropic subgroup $G \subset G' \subset A[n]$. As you can see, the vector space (or more accurately $\mathbb{Z}/n\mathbb{Z}$ module) is the $n$-torsion subgroup.
I can't think of a reference which could be considered "canonical" (maybe Mumford or Milne's books on abelian varieties), this is a standard definition and the adjective isotropic applies in some sufficiently general situation. For example, a free $R$-module $M$ of rank $n$ with a symplectic pairing $e \in \wedge^n M$ (this is the situation in your reference, though I guess they don't require their pairing to be alternating).
To explain how this works in the case in your reference, note that the polarisation the author (implicitly) equips $E \times E'$ with is the product polarisation ($E$ is naturally principally polarised). The upshot is that the pairing on $A[n]$ for $A = E \times E'$ is given by $$e_{A,n}((P,P'), (Q,Q')) = e_{E,n}(P,Q) e_{E',n}(P',Q')$$ for $P,Q \in E[n]$ and $P',Q' \in E'[n]$. The claim is then an exercise.