We need to count how many given species do not have an odd number of electrons, i.e. species having an even total number of electrons.

Count total electrons for each species using atomic numbers:

• $\text{N} = 7$
• $\text{O} = 8$
• $\text{I} = 53$
• $\text{Cl} = 17$
• $\text{Br} = 35$
• $\text{F} = 9$

Now evaluate each species:

(A) $\mathrm{NO_2}$ $7 + 2(8) = 7 + 16 = 23$ $23$ is odd, so it has odd electrons. Not counted.

(B) $\mathrm{ICl_4^-}$ $53 + 4(17) + 1 = 53 + 68 + 1 = 122$ ($+1$ because of the negative charge) $122$ is even, so it does not have odd electrons. Counted.

(C) $\mathrm{BrF_3}$ $35 + 3(9) = 35 + 27 = 62$ $62$ is even, so it does not have odd electrons. Counted.

(D) $\mathrm{ClO_2}$ $17 + 2(8) = 17 + 16 = 33$ $33$ is odd, so it has odd electrons. Not counted.

(E) $\mathrm{NO_2^+}$ $7 + 2(8) - 1 = 7 + 16 - 1 = 22$ ($-1$ because of the positive charge) $22$ is even, so it does not have odd electrons. Counted.

(F) $\mathrm{NO}$ $7 + 8 = 15$ $15$ is odd, so it has odd electrons. Not counted.

Count the species with even number of electrons:

• $\mathrm{ICl_4^-}$
• $\mathrm{BrF_3}$
• $\mathrm{NO_2^+}$

Total $= 3$

Therefore, the number of molecules/ions which do not have odd number of electrons is: $\boxed{3}$

Comparison with stored correct answer:

• Derived answer: $3$
• Stored correct answer: $3$
• They agree.