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Maybe I misunderstand something, but don't all curves have etale covers? Embed $X$ in $J^1$ (divisors of degree $1$ modulo linear equivalence). Then $J^1$ is a torsor for the Jacobian $J$ and since $J$ has etale covers, e.g. coming from multiplication by an arbitrary $n$, $J^1$ does too. Certainly, for those curves with a rational divisor of degree one, they have covers, as $J^1$ is isomorphic to $J$.

EDIT: Upon further reflection, I guess it's not true that $J^1$ always has covers, as it may not be in the divisible part of the Weil-Chatelet group of $J$. But there definitely exist curves with no points having divisors of degree one, and therefore covers of arbitrarily large degree.

However, you question is a good one and you might be heading in the direction of Grothendieck's section conjecture: For finitely generated fields $k$, $X(k)$ is non-empty if and only if there is a section $G_k \to \pi_1(X)$ of the canonical projection $\pi_1(X) \to G_k$, where $G_k$ is the absolute Galois group of $k$ and $\pi_1$ is the etale fundamental group.

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Maybe I misunderstand something, but don't all curves have etale covers? Embed $X$ in $J^1$ (divisors of degree $1$ modulo linear equivalence). Then $J^1$ is a torsor for the Jacobian $J$ and since $J$ has etale covers, e.g. coming from multiplication by an arbitrary $n$, $J^1$ does too. Certainly, for those curves with a rational divisor of degree one, they have covers, as $J^1$ is isomorphic to $J$.

However, you question is a good one and you might be heading in the direction of Grothendieck's section conjecture: For finitely generated fields $k$, $X(k)$ is non-empty if and only if there is a section $G_k \to \pi_1(X)$ of the canonical projection $\pi_1(X) \to G_k$, where $G_k$ is the absolute Galois group of $k$ and $\pi_1$ is the etale fundamental group.