1. Find total electrons in each species

   • For CO:

     • Carbon has $6$ electrons
     • Oxygen has $8$ electrons
     • Total $= 6+8 = 14$

   • For $NO^+$:

     • Nitrogen has $7$ electrons
     • Oxygen has $8$ electrons
     • Positive charge means one electron removed
     • Total $= 7+8-1 = 14$

   So, both CO and $NO^+$ are isoelectronic with $14$ electrons.

2. Use molecular orbital idea for 14-electron diatomic species

   Species with $14$ total electrons like $N_2$, CO, and $NO^+$ have the same valence MO filling pattern and hence the same bond order.

   For such species, $\text{Bond order} = \frac{N_b - N_a}{2}$ where $N_b$ = number of bonding electrons and $N_a$ = number of antibonding electrons.

3. Bond order of CO

   CO is isoelectronic with $N_2$, so its bond order is $3$

4. Bond order of $NO^+$

   $NO^+$ is also isoelectronic with $N_2$, so its bond order is $3$

5. Sum of bond orders

   $3 + 3 = 6$

Therefore, the required sum is $\boxed{6}$