The Hamiltonian is the energy, which is just one component of a four-vector and therefore not Lorentz invariant.
The Lagrangian is the Legendre transform of the Hamiltonian and I was wondering if there is some good reason why we get through the Legendre transform something invariant?
The previous answer is very good, but I think can be simplified a bit.
In particle mechanics, the Lagrangian $L$ is $$L = p\dot{q} - H$$ So let's look at this in special relativity. We get, with $p = \gamma m v$, $\dot{q} = v$, and $H = E = \gamma mc^{2}$, $$L = \gamma mv^{2} - \gamma mc^{2} = - \gamma mc^{2}(1 - (v/c)^{2}) = - mc^{2}/\gamma$$ It's not that $L$ is a scalar (which is what I thought originally), but that $\int L\,dt$ is a scalar. And this is easy, because $$\int L\,dt = -\int (mc^{2}/\gamma)\,dt = - \int mc^{2}\,d\tau$$ where $\tau$ is the proper time. This integral is clearly invariant, as we should wish for the action.
The Lagrangian is what is integrated over spacetime in the action, i.e. has to be a 4-form. As such, it is necessarily a (pseudo-)scalar under Lorentz transformations.
When wondering about Lorentz transformations and such, the Hamiltonian is, as a non-Lorentz-covariant object, not a good starting point, by the way. It is often better to start with the Lagrangian that makes the Lorentz covariance of the theory manifest.
We will here give our interpretation of OP's question (v4).
We assume that OP's definition of Lorentz covariance is that the equations of motion (eom) of the theory is Lorentz covariant.
We will assume that the theory has an action principle, and that the eoms are the Euler-Lagrange (EL) equations.
One may prove that Lorentz invariance of the action implies Lorentz covariance of the EL eqs., cf. e.g. this Phys.SE post.
The implication (3) does in principle not hold in the other direction, but in practice Lorentz covariant EL eqs. arise from a Lorentz invariant action principle.
Putting these facts together show that it is natural to expect that the action to be Lorentz invariant for a Lorentz covariant theory, cf. definition (1).
Next, we will assume that the Legendre transformation is well-defined.
Also we will assume that the Legendre transformation is an involution, i.e. performing the Legendre transformation twice gets us back to the starting point.
In particular, if OP starts from a Lorentz covariant (but not necessarily manifestly$^1$ Lorentz covariant) Hamiltonian formulation, this means that the Hamiltonian eoms are Lorentz covariant, cf. definition (1). The Hamiltonian$^2$ $H(q,p)$ itself is of course not Lorentz invariant, but the temporal component of a four-vector, as OP correctly writes. Points 2-4 now motivate that the Hamiltonian action $$S_H[q,p]~=~\int \!dt ~(p_i\dot{q}^i-H(q,p))$$ is Lorentz invariant. It follows that the Lagrangian action $S[q]$ is also Lorentz invariant.
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$^1$ For manifestly Lorentz covariant Hamiltonian formulations, see e.g. my Phys.SE answer here.
$^2$ The following argument can be extended to field theory.
The electric field may be expressed as a complex amplitude $\exp i(k \cdot x-\omega t)$, where $k=2\pi /\lambda$. The phase $k\cdot x-\omega t$ is Lorentz invariant. On the other hand, the phase is given by an integral over the path of the wave, i.e., $\int k \cdot \space dx -\omega t$. Now let us add the de Broglie's relations $p=\hbar k$ and $E=\hbar \omega$, thus we have $ \frac{1}{\hbar}\int_{x_a}^{x_b} p \cdot \space dx -\frac{1}{\hbar}E t= \frac{1}{\hbar}\int_{t_a}^{t_b} p \cdot \dot{x} \space dt - \frac{1}{\hbar}\int_{t_a}^{t_b} E \space dt = \frac{1}{\hbar}\int_{t_a}^{t_b} (2T-E) \space dt= \frac{1}{\hbar}\int_{t_a}^{t_b} L \space dt.$
