In a quasi-one-dimensional system (e.g., a tube) with low concentration of short-range defects the resistivity $\rho$ has peaks (Van Hove singularities), occurring when the Fermi energy $E_F$ crosses a bottom $E_N$ of some (resonant) subband of transversal quantization. Scattering smears these singularities in a very nontrivial manner. In the immediate vicinity of Van Hove singularity a plane wave is not a reasonable approximation for the electron wave functions in the resonant subband and one has to employ exact 1D results instead. Here the resistivity is dominated by scattering at rare ``twin'' pairs of close defects, while scattering at solitary defects is suppressed. The character of the relevant twin pairs crucially depends on the sign of the impurity potential (repulsive -- $\lambda>0$, or attractive -- $\lambda<0$).

In our previous paper we have studied the case of repulsing impurities, where the twins scattering has a nonresonant character: the main contribution to resistance comes from all pairs with separation $L\lesssim L_{\rm twin}\sim1/\lambda\ll \overline{L}$, $\overline{L}$ being the average distance between impurities.

In the present work we show that for attracting impurities the twins scattering is dominated by resonant twin pairs with certain sharply defined separations $L\approx L_{\rm twin}(\varepsilon)$, depending on $\varepsilon\equiv E_F-E_N$. The resonant separation $L_{\rm twin}(\varepsilon)$ is determined from the equation $E_{\rm twin}(L)=\varepsilon$, where $E_{\rm twin}(L)$ is the bound state energy for a system of two impurities with separation $L$. It is important that for $\varepsilon<0$ the above equation has solutions only in the attracting case. The resistivity in the case of attractive impurities is larger than for repulsive case due to resonant character of scattering.