Solution:

Let the number be $n$ with digits $d_1d_2\ldots d_{17}$, so that the reversed number $m$ has digits $d_{17}d_{16}\ldots d_1$. Let the digits of $n + m$ be $a_0a_1\ldots a_{17}$, where $a_0$ may be zero. Let the carry forward when adding digits to get $a_i$ be $c_{i - 1}$, so that, in general, $c_i + d_i + d_{18 - i} = a_i + 10 c_{i - 1}$. Obviously $c_i$ is $0$ or $1$.

Suppose all the digits $a_i$ are odd (except that $a_0$ may be zero). Now $c_9 + 2d_9 = a_9 + 10 c_8$. Since $a_9$ is odd, $c_9$ must be $1$. But if we consider $c_{10} + d_{10} + d_8 = a_{10} + 10 c_9$, we see that since $a_{10}$ is odd it is at least $1$ and hence $d_8 + d_{10}$ is at least $10$. Hence there must be a non-zero carry $c_9$ in $c_{10} + d_{10} + d_8 = a_{10} + 10 c_9$ irrespective of the value of $c_{10}$.

We can now iterate and conclude successively that $c_{12}$, $c_{14}$, $c_{16}$ must be non-zero.