Which of the following species is paramagnetic?

The magnetic nature of a species can be determined by its molecular orbital electronic configuration.

For $\text{N}_2$ (14 electrons): \sigma_{1s}^{2} \sigma^{*}_{1s}^{2} \sigma_{2s}^{2} \sigma^{*}_{2s}^{2} \pi_{2p_x}^{2} = \pi_{2p_y}^{2} \sigma_{2p_z}^{2}. All electrons are paired, so it is diamagnetic.

For $\text{Li}_2$ (6 electrons): \sigma_{1s}^{2} \sigma^{*}_{1s}^{2} \sigma_{2s}^{2}. All electrons are paired, so it is diamagnetic.

For $\text{F}_2$ (18 electrons): \sigma_{1s}^{2} \sigma^{*}_{1s}^{2} \sigma_{2s}^{2} \sigma^{*}_{2s}^{2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} = \pi_{2p_y}^{2} \pi^{*}_{2p_x}^{2} = \pi^{*}_{2p_y}^{2}. All electrons are paired, so it is diamagnetic.

For $\text{N}_2^{+}$ (13 electrons): \sigma_{1s}^{2} \sigma^{*}_{1s}^{2} \sigma_{2s}^{2} \sigma^{*}_{2s}^{2} \pi_{2p_x}^{2} = \pi_{2p_y}^{2} \sigma_{2p_z}^{1}. It has one unpaired electron, so it is paramagnetic.

Answer: $\text{N}_2^{+}$