One of the possible versions of quantum mechanics, known as Kuryshkin-Wodkiewicz quantum mechanics, is considered. In this version, the quantum distribution function is positive, but, as a retribution for this, the von Neumann quantization rule is replaced by a more complicated rule, in which an observed value $A$ is associated with a pseudodifferential operator ${\hat{O}(A)}$. This version is an example of a dissipative quantum system and, therefore, it was expected that the eigenvalues of the Hamiltonian should have imaginary parts. However, the discrete spectrum of the Hamiltonian of a hydrogen-like atom in this theory turned out to be real-valued. In this paper, we propose the following explanation for this paradox. It is traditionally assumed that in some state ${\psi}$ the quantity $A$ is equal to ${\lambda}$ if ${\psi}$ is an eigenfunction of the operator ${\hat{O}(A)}$. In this case, the variance ${\hat{O}((A-\lambda)2)\psi}$ is zero in the standard version of quantum mechanics, but nonzero in Kuryshkin’s mechanics. Therefore, it is possible to consider such a range of values and states corresponding to them for which the variance ${\hat{O}((A-\lambda)2)}$ is zero. The spectrum of the quadratic pencil ${\hat{O}(A2)-2\hat{O}(A)\lambda + \lambda 2 \hat{E}}$ is studied by the methods of perturbation theory under the assumption of small variance ${\hat{D}(A) = \hat{O}(A2) - \hat{O}(A) 2}$ of the observable $A$. It is shown that in the neighborhood of the real eigenvalue ${\lambda}$ of the operator  ${\hat{O}(A)}$, there are two eigenvalues of the operator pencil, which differ in the first order of perturbation theory by  ${\pm i \sqrt{\langle \hat{D} \rangle}}$.