= Solution
Because the present $\gamma_0$ is anti-Hermitian, choose the <Dirac adjoint> with Hermitian intertwining matrix $A=i\gamma_0$:
$$
\bar\psi=\psi^\dagger A,\qquad A=i\gamma_0=\begin{pmatrix}0&-i\\i&0\end{pmatrix}.
$$
This convention makes the <Dirac scalar bilinear> $\bar\psi\psi$ Hermitian. For real $m$, a <Lorentz invariant> <mass term> is
$$
\boxed{\mathcal L_m=-m\bar\psi\psi
=im\left(\psi_+^*\psi_- -\psi_-^*\psi_+\right).}
$$
To check infinitesimal <Lorentz invariance>, put $K=\sigma_3/2$. Then $K^\dagger=K$ and $K^\dagger A+AK=\{K,A\}=0$. Using <3/a>,
$$
\delta\mathcal L_m=-m\theta\psi^\dagger(K^\dagger A+AK)\psi=0.
$$
Equivalently, $\delta(\psi_+^*\psi_-)=\theta\psi_+^*\psi_-/2-\theta\psi_+^*\psi_-/2=0$, and similarly for its <complex conjugate>. Exponentiation gives $S^\dagger AS=A$ and therefore $\bar\psi'(x')=\bar\psi(x)S^{-1}$.
At fixed coordinates $\mathcal L_m$ has only the orbital variation of a <Lorentz scalar>; the spacetime measure is invariant under the connected <Lorentz group>, so the integrated <action> is invariant. The <mass term> pairs the two opposite <chiralities>; retaining only one <Weyl spinor> removes this Dirac mass. An equivalent spinor phase convention gives a real multiple of $\psi_+^*\psi_-+\psi_-^*\psi_+$ instead.
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