Let $Y$ be an algebraic curve of genus $g \geq 1$ defined over a number field $K$. If $Y$ has a $K$-point, then one can define the Abel-Jacobi map which embeds $Y$ into its Jacobian variety $\text{Jac}(Y)$. Torelli's theorem says that, conversely, $\text{Jac}(Y)$ essentially determines the curve $Y$ (this is an exact statement over an algebraically closed field).
The issue is that many (perhaps most... a theorem of Bhargava states concretely that most, in terms of natural density, of hyperelliptic curves of large genus defined over $\mathbb{Q}$ do not have any $\mathbb{Q}$-points) algebraic curves of genus $g \geq 1$ do not have $K$-rational points at all. Nevertheless, their Jacobian can still be defined over $K$.
For example, many genus one curves given by $C_f: z^2 = f(x,y)$, for $f$ a binary quartic form defined over $\mathbb{Q}$, do not have rational points. Indeed, even when $C_f$ is everywhere locally soluble, it may still fail to have $\mathbb{Q}$-rational points. Nevertheless it is clear that its Jacobian $\text{Jac}(C_f)$ is an elliptic curve defined over $\mathbb{Q}$, hence has at least one rational point. Moreover every elliptic curve is isomorphic to its Jacobian, so every Jacobian abelian variety of genus 1 is the Jacobian of a genus one curve with a rational point.
My question is: do there exist Jacobian abelian varieties $A$ defined over a number field $K$, of dimension $g > 1$, such that for all algebraic curves $Y$ defined over $K$ with $\text{Jac}(Y)$ is isomorphic to $A$ over $K$, $Y$ has no $K$-rational point?
