Let $S = Spec(O_K)$ be the spectrum of the rings of integers of a number field $K$. Let $A/S \setminus T$ be an Abelian scheme over an open subscheme $S \setminus T \subseteq S$. Does the kernel of $n$multiplication $A[n]$ become constant after a (non empty) étale base change $S' \to S \setminus \{x_1,\ldots,x_n\}$?
2 Answers
Do you really mean to consider an abelian scheme over the entire ring of integers, and not just a localization thereof? Either way, every finite flat group scheme $G$ over a domain $R$ with fraction field $F$ such that ${\rm{char}}(F)$ does not divide the order $n$ of $G$ is etale over $R[1/n]$ (as can be checked on geometric fibers, using that the identity component on such fibers has order which divides $n$ yet is a multiple of the residue characteristic, forcing the identity component to be trivial and hence the geometric fiber to be etale at the identity and therefore etale everywhere via translation). Thus, as for any finite etale scheme with constant fiber rank over any base at all, it becomes constant over a finite etale cover of $R[1/n]$ (using base change against itself to split off the diagonal and then induct on fiberdegree).

$\begingroup$ "using base change against itself to split off the diagonal" Do you mean this: $X/S$ is finite étale. $\Delta: X \to X \times_S X/X$ is finite étale and a closed immersion, so an open and closed immersion, so $((X \times_S X)  X)/S$ is again finite étale, now use induction? $\endgroup$– user5262Commented Aug 10, 2014 at 14:56

In less fancy language, you are asking for the ramification of the extension of $K$ where you adjoin the coordinates of the $n$torsion points of $A$. Under your assumption of everywhere good reduction, only places dividing $n$ can ramify. So there may be more or less than $n$ points $x_i$ that you have to remove from $S$.