Q:

Number consists of 17 digits, other number is first number in reverse order. Prove that their sum has at least one even digit

My thought was that it doesn't matter, whatever is it 17 digits in this problem, or any other odd number.
So for simplicity we can think of 3 digits number. 
k1 k2 k3
k3 k2 k1

Points that $\lceil k/2 \rceil$ should be even. 
However, it is true only in case when there is not such pair, which sum is bigger than 9.

A:

Let the digits be $d_1$ through $d_{17}$, and suppose that every digit in the sum is odd. The addition is shown below. In the middle we have $d_9+d_9$, which will yield an even digit unless there is a carry from the column to the right, $d_8+d_{10}$; I’ve indicated this carry with a $c$ below the line. This means that the column to the left, $d_{10}+d_8$, must produce a carry to be added to the $d_{11}+d_7$ column to its left; this is again indicated by a $c$ below the line.
In order for $d_{11}+d_7$ plus the carry to yield an odd digit, $d_{11}+d_7$ must yield an even digit. This means that there must be a carry into the $d_7+d_{11}$ column from the $d_6+d_{12}$ column, and again I’ve indicated it with a $c$ below the line. Since $d_6+d_{12}$ produces a carry, so does $d_{12}+d_6$, and there is a carry in the $d_{13}+d_5$ column. This means that $d_5+d_{13}$ by itself yields an even digit, so there must be a carry into that column from the $d_4+d_{14}$ column. The $d_{14}+d_4$ column then produces a carry into the $d_{15}+d_3$ column, so $d_3+d_{15}$ yields an even digit, and there must be a carry from the $d_2+d_{16}$ column. The $d_{16}+d_2$ column therefore produces a carry into the $d_{17}+d_1$ column, completing the table.
$$\begin{array}{ccc}
d_{17}&d_{16}&d_{15}&d_{14}&d_{13}&d_{12}&d_{11}&d_{10}&d_9&d_8&d_7&d_6&d_5&d_4&d_3&d_2&d_1\\
d_1&d_2&d_3&d_4&d_5&d_6&d_7&d_8&d_9&d_{10}&d_{11}&d_{12}&d_{13}&d_{14}&d_{15}&d_{16}&d_{17}\\ \hline
c&&c&&c&&c&&c&&c&&c&&c
\end{array}$$
But now we have a contradiction: $d_1+d_{17}$ must yield an odd digit at the righthand end of the sum, so $d_{17}+d_1$ plus the carry must yield an even digit in the $17$-th column from the right.

