When does $V=L$ becomes inconsistent?

The borderline seems to be very near the $\omega_1$-Erdős cardinals.

The Wikipedia page on $0^\sharp$ explains it thus:

The existence of $\omega_1$-Erdős cardinals implies the existence of $0^\sharp$. This is close to being best possible, because the existence of $0^\sharp$ implies that in the constructible universe there is an $\alpha$-Erdős cardinal for all countable $\alpha$, so such cardinals cannot be used to prove the existence of $0^\sharp$.

In particular, an $\omega_1$-Erdős cardinal is inconsistent with $V=L$, but having $\alpha$-Erdős cardinals for countable $\alpha$ is fine with $V=L$. All the smaller large cardinals in the hierarchy (see the Wikipedia large cardinal list) also appear to relativize down to $L$.

Just a supplement to Joel's answer:

The exact boundary for existence of $0^\sharp$ was pinned down by Klaus Gloede: $0^\sharp$ exists if and only if there is a cardinal $\kappa$ such that every constructible partition of $[\kappa]^{<\omega}$ has an uncountable homogeneous set.

Gloede, Klaus, Ordinals with partition properties and the constructible hierarchy. Z. Math. Logik Grundlagen Math. 18 (1972), 135–164.

As Joel points out, everything below this on the usual large cardinal hierarchy seems to be compatible with L.