What conservation law corresponds to Lorentz boosts?

Warning: this is a long and boring derivation. If you are interested only in the result skip to the very last sentence.

Noether's theorem can be formulated in many ways. For the purposes of your question we can comfortably use the special relativistic Lagrangian formulation of a scalar field. So, suppose we are given an action $$S[\phi] = \int {\mathcal L}(\phi(x), \partial_{\mu} \phi(x), \dots) {\rm d}^4x.$$

Now suppose the action is invariant under some infinitesimal transformation $m: x^{\mu} \mapsto x^{\mu} + \delta x^{\mu} = x^{\mu} + \epsilon a^{\mu}$ (we won't consider any explicit transformation of the fields themselves). Then we get a conserved current $$J^{\mu} = {\partial {\mathcal L} \over \partial \phi_{,\mu}} \phi^{,\nu} a_{\nu} - {\mathcal L} a^{\mu} = \left ({\partial {\mathcal L} \over \partial \phi_{,\mu}} \phi^{,\nu} - {\mathcal L} g^{\mu \nu} \right) a_{\nu} .$$ We obtain a conserved charge from it by letting $Q \equiv \int J^0 {\rm d}^3x$ since from $\partial_{\mu}J^{\mu} =0$ we have that $$ {\partial Q \over \partial t} = \int {\rm Div}{\mathbf J}\, {\rm d}^3 x = 0$$ which holds any time the currents decay sufficiently quickly.

If the transformation is given by translation $m_{\nu} \leftrightarrow \delta x^{\mu} = \epsilon \delta^{\mu}_{\nu}$ we get four conserved currents $$J^{\mu \nu} = {\partial {\mathcal L} \over \partial \phi_{\mu}} \phi^{\nu} - {\mathcal L} g^{\mu \nu} .$$

This object is more commonly known as stress energy tensor $T^{\mu \nu}$ and the associated conserved currents are known as momenta $p^{\nu}$. Also, in general the conserved current is simply given by $J^{\mu} = T^{\mu \nu} a_{\nu}$.

For a Lorentz transformation we have $$m_{\sigma \tau} \leftrightarrow \delta x^{\mu} = \epsilon \left(g^{\mu \sigma} x^{\tau} - g^{\mu \tau} x^{\sigma} \right)$$ (notice that this is antisymmetric and so there are just 6 independent parameters of the transformation) and so the conserved currents are the angular momentum currents $$M^{\sigma \tau \mu} = x^{\tau}T^{\mu \sigma} - x^{\sigma}T^{\mu \tau}.$$ Finally, we obtain the conserved angular momentum as $$M^{\sigma \tau} = \int \left(x^{\tau}T^{0 \sigma} - x^{\sigma}T^{0 \tau} \right) {\rm d}^3 x . $$

Note that for particles we can proceed a little further since their associated momenta and angular momenta are not given by an integral. Therefore we have simply that $p^{\mu} = T^{\mu 0}$ and $M^{\mu \nu} = x^{\mu} p^{\nu} - x^{\nu} p^{\mu}$. The rotation part of this (written in the form of the usual pseudovector) is $${\mathbf L}_i = {1 \over 2}\epsilon_{ijk} M^{jk} = ({\mathbf x} \times {\mathbf p})_i$$ while for the boost part we get $$M^{0 i} = \left(t {\mathbf p} - {\mathbf x} E \right)^i $$ which is nothing else than the center of mass at $t=0$ (we are free to choose $t$ since the quantity is conserved) multiplied by $\gamma$ since we have the relations $E = \gamma m$, ${\mathbf p} = \gamma m {\mathbf v}$. Note the similarity to the ${\mathbf E}$, $\mathbf B$ decomposition of the electromagnetic field tensor $F^{\mu \nu}$.

To supplement Marek's execllent answer, I provide an alternative derivation below and provide as many intermediate steps as possible.

For an infinitesimal displacement $y^\mu=x^\mu+\xi^\mu$, a scalar field changes as

$$\phi(y)=\phi(x)+\xi^\mu \partial_\mu\phi(x)+...$$

The displacement by infinitesimal Lorentz transform $\Lambda^{\mu\nu}$ is $y^\mu=x^\mu+\Lambda^{\mu\nu}x_\nu$. Similarly the scalar field changes as: $$\phi(y)=\phi(x)+ \Lambda^{\mu\nu}x_\nu\partial_\mu\phi(x)+...$$ The variation of the field w.r.t. $\Lambda^{\mu\nu}$ is $$\frac{\delta \phi}{\delta \Lambda^{\mu\nu}}=x_\nu\partial_\mu\phi(x)-x_\mu\partial_\nu\phi(x)$$ The reason there are two terms on the right hand side is because infinitesimal Lorentz transform $\Lambda^{\mu\nu}$ is anti-symmetric, i.e. $\Lambda^{\nu\mu} = -\Lambda^{\mu\nu}$, which has only 6 independent components. (You can verify this by demanding the scalar product is unchanged after transformation, $y^\mu y_\mu = x^\mu x_\mu$)

Using Principle of Least Action, variation in Lagrangian $\mathcal{L}$ is

$$\frac{\delta \mathcal{L}}{\delta \Lambda^{\mu\nu}}=\sum_n\{\frac{\partial \mathcal{L}}{\partial\phi_n} \frac{\delta\phi_n}{\delta \Lambda^{\mu\nu}} + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi_n)} \frac{\delta(\partial_{\mu}\phi_n)}{\delta \Lambda^{\mu\nu}} \}$$ Applying the equation of motion $$\frac{\partial \mathcal{L}}{\partial \phi_n} -\partial_\mu\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}=0$$ we get the conservation law: $$\frac{\delta \mathcal{L}}{\delta \Lambda^{\mu\nu}}=\sum_n\partial_\mu[\frac{\partial \mathcal{L}}{\partial_{\mu}\phi_n} \frac{\delta\phi}{\delta \Lambda^{\mu\nu}} ] $$ Substituting the expression for $\delta \phi/\delta \Lambda^{\mu\nu}$ and a similar one for $\delta \mathcal{L}/\delta \Lambda^{\mu\nu}$, we get the final conservation law $$\partial_\mu j^{\mu \lambda\sigma} = 0 $$ where the conservative current $$j^{\mu \lambda\sigma}=x^\lambda T^{\mu\sigma} - x^{\sigma}T^{\mu\lambda}$$ is the angular momentum and $$T_{\mu\nu}= \frac{\partial\mathcal{L}}{\partial(\partial_\mu \phi)}\partial_\nu\phi-g_{\mu\nu}\mathcal{L} $$ is the momentum.