The $\ce{N2^+}$ molecule has the molecular orbitals $$\ce{\sigma(1s)^2\; \sigma^*(1s)^2\; \sigma(2s)^2\; \sigma^*(2s)^2\; \pi(2p_x)^2 \; \pi(2p_y)^2 \; \sigma(2p_z)^1,}$$ which can be seen on the following picture:
So, the ground state has $\mathrm{A_g}$ symmetry in $D_\mathrm{2h}$ point group, i.e. $\Sigma_\mathrm{g}^+$ in $D_\mathrm{\infty h}$.
But what will the excited states look like?
I'd expect, that the first excited electrons will be the ones, which "need" the smallest amount of energy to get into a higher state.
So, the first electron will be excited from $\ce{\pi(2p_x)}$ or $\ce{\pi(2p_y)}$ orbitals to $\ce{\sigma(2p_z)}$ like this:
The first excited state will have $\mathrm{B_{3u}}$ or $\mathrm{B_{2u}}$ symmetry in $D_\mathrm{2h}$.
The second excited state would be
I.e. it would have the $\mathrm{B_{3g}}$ or $\mathrm{B_{2g}}$ symmetry.
Is my expectation correct or will the electrons be excited in a different order? And what will the 3rd and 4th excited states look like?