5. Reverse the order of the three digits of a three-digit number, and add the resulting number to the original number. If none of the digits in the sum is even, then the number is called a "odd-sum number". Therefore, among all three-digit numbers, the number of odd-sum numbers is ( ).
(A) 100
(B) 120
(C) 160
(D) 200
5.A.

Let the three-digit number be $\overline{a_{1} a_{2} a_{3}}$. Then
$$
\begin{array}{l}
\overline{a_{1} a_{2} a_{3}}+\overline{a_{3} a_{2} a_{1}} \\
=100\left(a_{1}+a_{3}\right)+10\left(a_{2}+a_{2}\right)+\left(a_{1}+a_{3}\right)
\end{array}
$$

If $a_{1}+a_{3}$ does not carry over, then the tens digit of the sum must be even, which does not meet the requirement. Therefore, $a_{1}+a_{3}=11,13,15,17$.

Since $11=9+2=8+3=7+4=6+5$, then $a_{1} 、 a_{3}$ have $4 \mathrm{~A}_{2}^{2}$ possible values;

Since $13=9+4=8+5=7+6$, then $a_{1} 、 a_{3}$ have $3 \mathrm{~A}_{2}^{2}$ possible values;

Since $15=9+6=8+7$, then $a_{1} 、 a_{3}$ have $2 \mathrm{~A}_{2}^{2}$ possible values;
Since $17=9+8$, then $a_{1} 、 a_{3}$ have $\mathrm{A}_{2}^{2}$ possible values.
Since $a_{2}+a_{2}$ cannot carry over, $a_{2}$ can only take $0,1,2,3,4$.
Therefore, the number of numbers that meet the condition is
$$
5\left(4 \mathrm{~A}_{2}^{2}+3 \mathrm{~A}_{2}^{2}+2 \mathrm{~A}_{2}^{2}+\mathrm{A}_{2}^{2}\right)=100 \text { (numbers) }
$$