Can eigenvalues of the density matrix in the Lindblad equation be negative? Can the density matrix in the Lindblad equation for an open mixed quantum system have (real part) negative eigenvalues?
 A: The density matrix by definition is a positive semi-definite matrix. It therefore has real positive eigenvalues.
Physically, these eigenvalues correspond to the probability $p_i$ of being in one of the pure eigenstates $|\phi_i\rangle$ according to
$$\rho = \sum_i p_i |\phi_i\rangle\langle \phi_i|.$$
Clarification in response to the comments
The above answer only uses the definition of the density matrix. Positive semi-definiteness of $\rho$ is a very basic property which follows from its probability interpretation.
The above also allows for mixed states. As @ZeroTheHero states, a pure state would be the case where one of the $p_i=1$, such that $\rho=|\phi_i\rangle\langle \phi_i|$.
According to the comments, the OP may contain a second question, which is related to the dynamics generated by the Lindblad equation. Indeed, the Lindblad equation manifestly conserves positive semi-definiteness of the density matrix. It is also trace preserving, since another important property of the density matrix is that the probabilities $p_i$ sum to 1.
Note that the Lindblad equation is particularly nice because of this property. While it usually is derived using multiple approximations, it produces a fully consistent density matrix at all times of the dynamics. This is not the case for many other ways to treat open system dynamics. E.g. the Redfield equation contains less approximations, but does not generate positive time evolution.
