National Math Olympiad

Write the three-digit number as $abc$. The number we obtain by reversing the order of the digits is $cba$. The sum $x$ of these two numbers is $$x=\overline{abc}+\overline{cba}=(a+c)10^2+(b+b)10+(c+a).$$ Since all the digits of $x$ are odd and $2b$ is even, we conclude that $c+a$ is equal to at least 10. The number $c+a$ cannot be equal to 10 because the units of $x$ are odd and equal to the units of $c+a$. Hence, $c+a$ is at least 11 and $a$ is at least 2. The number $abc$ will be the smallest when we choose the smallest possible $a$, i.e. $a=2$. In this case $c=9$ and the smallest possible value of $b$ is

We see that the number $209+902=1111$ consists of only odd digits. Thus, $209$ is the smallest number with the desired properties.