Step 1: Recall the bond order formula.
In molecular orbital theory, bond order measures how strongly atoms are joined. It is found from \[ \text{Bond order} = \frac{N_b - N_a}{2} \] where $N_b$ is the number of electrons in bonding orbitals and $N_a$ is the number in antibonding orbitals.
Step 2: Link paramagnetism to unpaired electrons.
A species is paramagnetic when it has one or more unpaired electrons. If all electrons are paired, it is diamagnetic. So we need a species with both bond order $2.5$ and at least one unpaired electron.
Step 3: Start from neutral nitrogen.
Neutral $N_2$ has a bond order of $3$ and is diamagnetic, since all its electrons are paired. To change the bond order we must add or remove an electron.
Step 4: Remove one electron to form $N_2^+$.
Taking out one electron from a bonding orbital of $N_2$ lowers the bond order by $0.5$: \[ 3 - 0.5 = 2.5 \] So $N_2^+$ has the required bond order of $2.5$.
Step 5: Check the magnetic nature of $N_2^+$.
Removing one electron leaves an odd number of electrons, so one orbital now holds a single unpaired electron. This makes $N_2^+$ paramagnetic.
Step 6: Compare with the other choices.
$O_2$ has bond order $2$, $O_2^{2-}$ has bond order $1$, and $C_2$ has bond order $2$ and is diamagnetic. None of these fit, so the answer is $N_2^+$.
\[ \boxed{N_2^{+}} \]