Suppose that $C\_1, C\_2$ are two curves of genus $g \geq 2$ defined over a number field $K$. Let $J\_1, J\_2$ respectively be their Jacobians. Suppose that $J\_1, J\_2$ are isogenous over $K$ and $C\_1(K), C\_2(K)$ are both non-empty, can $C\_1(K), C\_2(K)$ have different cardinalities?


For $g = 1$ and without the assumption that $C\_i(K) \ne \emptyset$, the conclusion is obviously false. Take any elliptic curve $E$ over $\mathbb{Q}$ of positive rank such that the $2$-Selmer group of $E$ is non-trivial, so that there is a genus one curve $C$ which represents a non-trivial $2$-Selmer element of $E$. Then $E$ is isomorphic to the Jacobian of $C$, and the Jacobian of $E$ is itself, so that $C,E$ have isogenous Jacobians but $E(\mathbb{Q})$ is by assumption infinite but $C(\mathbb{Q}) = \emptyset$. 



Curves with isogenous Jacobians
Yes it is possible, already for $(K,g) = ({\bf Q},2)$, and
already with the first example of isogenous $C\_1,C\_2$ listed in the LMFDB:
curve
[249.a.249.1](http://www.lmfdb.org/Genus2Curve/Q/249/a/249/1),
$y^2 + (x^3+1) y = x^2 + x$, has one rational Weierstrass point, while curve
[249.a.6723.1](http://www.lmfdb.org/Genus2Curve/Q/249/a/6723/1),
$y^2 + (x^3+1) y = -x^5 + x^3 + x^2 + 3x + 2$, has two, so the rational-point
counts don't even have the same parity. (The counts are known to be
$5$ and $6$ respectively.) Further examples are isogeny class
[277.a](http://www.lmfdb.org/Genus2Curve/Q/277/a)
(two curves, each with a unique rational Weierstrass point,
but the total counts are $1$ and $5$), and isogeny class
[644.a](http://www.lmfdb.org/Genus2Curve/Q/644/a)
(two curves, one with two rational points, the other not solvable over $\bf R$
--- so this violates your assumption that each $C\_i(K)$ is nonempty).
The first example where the counts *do* agree is the two curves in
isogeny class [294.a](http://www.lmfdb.org/Genus2Curve/Q/294/a),
each with $4$ rational points.


By the way, for $g=1$ there are examples even with each $C\_i(K)$ nonempty;
for example, take $K=\bf Q$, let $C\_1$ be either $X\_0(11)$ or $X\_1(11)$,
and let $C\_2$ be the third elliptic curve of conductor $11$.
Then $C\_1$ has five rational points and $C\_2$ has only one.


