Solution
= Solution
If every point of $E[n]$ is rational, the <Weil pairing>
$$
e_n:E[n]\times E[n]\longrightarrow\mu_n
$$
and its nondegeneracy imply that every $n$th root of unity belongs to $\mathbb Q$. The only roots of unity in $\mathbb Q$ are $\pm1$, so $n\leq2$. Since the question assumes $n\geq2$, only $n=2$ is possible. It does occur: any nonsingular equation
$$
y^2=(x-a)(x-b)(x-c)
$$
with distinct $a,b,c\in\mathbb Q$ has all four points of $E[2]$ rational.