Is $N_R$ a Majorana field in the Seesaw Lagrangian? Consider the Lagrangian for the type-I seesaw given by
$$-\mathcal{L}=\bar{\nu}_{L}m_DN_{R}+\frac{1}{2}\overline{(N_{R})^c}M_R N_{R}+\text{h.c.}.$$ 
$\bullet$ In this Lagrangian, what is the nature of $N_R$ field? I think it cannot be Majorana even though we write a Majorana mass for it because we don't take $(N_R)^c=N_R$ (the defining condition for a field to be Majorana).
$\bullet$ I know that after diagonalization, the light, and heavy neutrinos will be linear combinations of the fields $\nu_L$ and $N_R$, and are all  Majorana particles. But my question is what can we say about the fields $N_R$? 
 A: A Dirac spinor can be written as the sum of two Weyl (chiral) spinors, eigenstates of $\gamma_5$,
$$
\psi= \psi_L + \psi_R,\tag{1}
$$
but also, alternatively, as the sum of two Majorana (self-conjugate) spinors, eigenstates of C (and $i\gamma_2$)
$$ 
\psi= \chi + i\omega= \chi^c + i\omega^c,\tag{2}
$$
where each Majorana component is real in the Majorana representation, in which 
$\gamma_5= \sigma^3\otimes \sigma^2$, imaginary, $i\gamma_2= \sigma^2\otimes\sigma^2$, real, and $C=-i\sigma^1\otimes \sigma^2$, real. Note C and $i\gamma_2$ do not commute with $\gamma_5$ (a feature of 4 dimensions).
Consequently, a Majorana spinor cannot be Weyl, and a Weyl spinor cannot be Majorana. The Majorana and Weyl bases are mutually exclusive bases for the components of a Dirac spinor. 
$N_R$ is a Weyl spinor, and combines with $\nu_L$ to resolve into two Majorana components. Don't worry about N vs ν, they are meant to recall EW quantum numbers.

A comprehensive treatment is in section 13.2 of ISBN-13: 978-0198506218,
Gauge Theory of Elementary Particle Physics: Problems and Solutions, 1st Edition
by Ta-Pei Cheng &  Ling-Fong Li.
Edit as per references request : M Schwartz's notes, authoritative alright,  illustrate the link in the Weyl, not Majorana basis, where everything is complex, instead, so contrast eqns (9) to (34). 
A: The answer sits in the definition of charge conjugation as: 
$$\psi^c = C \bar{\psi}^T$$
where $C$ is some ''book dependent matrix''. In many notes, it is chosen to be $i\gamma_2$ but this will be irrelevant for our discussion.
With this in mind, we can now rewrite the action from your question as:
$$-\mathcal{L} = \bar{\nu_L} m_D N_R + \frac{1}{2}C (N_R)^T M_R N_R + h.c.$$
At which point is becomes very clear that $N_R$ is indeed a Majorana neutrino. 
