Why is the real part of optical conductivity dissipative although the real part of susceptibility is dispersive?

I understand your confusion between the apparent opposite roles of the optical conductivity, $sigma$, and the electric susceptibility, $chi$. Both are response functions: $$ P(t)=epsilon_0int_{-infty}^t chi(t-t’)E(t’)dt’$$ $$ J(t)=int_{-infty}^t sigma(t-t’)E(t’)dt’$$ where $P$ is the polarization density, and $J$ is the current density.

Their difference comes from how $P$ and $J$ interact with an electromagnetic wave. Check out Ampere’s Law: $$nablatimes H=J+frac{partial D}{partial t}.$$

Let’s replace the source terms with the Fourier transforms of the response functions (and using $D=epsilon_0E+P=epsilon E$): $$nablatimes H=sigma E+frac{partial epsilon E}{partial t}.$$

For a monochromatic wave, a time derivative amounts to multiplying by $iomega$. So we have $$nablatimes H=(sigma+iomegaepsilon) E.$$

So there you have it! $E$ generates $H$ in an electromagnetic wave through $sigma$ and $epsilon$. They play exactly the same role in Ampere’s Law, except due to the time derivative in Maxwell’s addition, $epsilon$ has an extra factor of $i$. Voila! The imaginary part of $sigma$ acts like the real part of $epsilon$ (or $chi$).