Langevin equation. What is the meaning of temperature? Consider a system of $N$ particles, subject to some interaction potential $U$ (e.g. Lennard-Jones) and to thermal noise. The equation of motion is given using the Langevin equation:
$$m_i \ddot{\bf r}_i = -\nabla U({\bf r}_i) -\lambda\dot{\bf r}_i + {\boldsymbol \eta}_i,$$
where $\lambda$ is the friction coefficient and ${\boldsymbol \eta} = [\eta_{i,x} ~ \eta_{i,y} ~\eta_{i,z}]$ is a vector of random gaussian forces, with $0$ mean and
$$\langle \eta_{i,\alpha}(t), \eta_{j,\beta}(t') \rangle = 2\lambda k_BT \delta_{i,j}\delta_{\alpha, \beta}\delta(t-t').$$
It seems (and maybe I'm wrong!!!) that $T$ is the temperature of the system. Well, I have difficulties to understand this point, since we are assuming that $T$ is a constant parameter (i.e. a thermal bath), but the system evolves over time, gaining (or loosing) kinetic energy, and therefore the temperature is changing over time as well!!!
Indeed,
$$T = \frac{1}{3Nk_B}\sum_{i=1}^N m_i \dot{\bf r}_i^2$$
changes over time according to the evolution of the velocity $\dot{\bf r}_i$.
What am I missing about the definition of this system and the role of the temperature?
 A: You are completely correct, that $T$ is a fixed parameter that defines the temperature of the system. The mistake is identifying this with the instantaneous "temperature" defined via your equation involving velocities. This equation is only true if you ensemble-average the right hand side, the appropriate ensemble being the canonical ensemble. Incidentally, the Langevin equation can be proven to generate states (coordinates and momenta) sampled from the canonical ensemble at temperature $T$. So,
$$T = \left\langle \frac{1}{3N}\sum_{i=1}^N m_i \dot{\bf r}_i^2\right\rangle$$
You should think of this equation not as a definition of $T$, but as a statement of what the ensemble average on the right turns out to be,
given that the system is at equilibrium at temperature $T$. The same equation applies to each individual atom $i$, if we omit the summation over $i$ and the normalising factor $N$.
Alternatively, if you consider a single realisation of the Langevin equation, you can consider time-averaging the right hand side of that equation and (subject to some reasonable assumptions about ensemble equivalence, ergodicity and so on) the result will also be $T$. 
The time dependent analogue of $T$, which you can call $T(t)$ if you like, although I prefer to emphasise the difference by calling it $\mathcal{T}(t)$ or something similar, is a mechanical variable: a function of coordinates and momenta in general (in this case, just momenta, or velocities). It fluctuates by definition, and is only equal to $T$ after averaging, and in an equilibrium ensemble. You could, with equal validity, define an instantaneous "temperature" for each atom $i$, using a similar formula, omitting the sum and the $1/N$, all fluctuating and all different. However they would all give the same ensemble average.
This distinction is not unique to the Langevin equation. For example, in molecular dynamics using the completely deterministic Nosé-Hoover thermostat, $T$ is again a fixed parameter, and it is possible to define an instantaneous "temperature" from the kinetic energy, with the same equation (more or less) as you have given, whose average is $T$.
