Parity of the digits present in the sum of a number and its reverse. Let $S_N$ denote sum of a $N$-digit number and its reverse,then prove;


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*If $N$ is of the form $4k+1$, at least one of the digits of $S_N$  is an even number. 

*If $N$ is of the form  $4k+3$, there exist at least one $N$-digit number whose $S_N$  contains only odd digits.
 A: Let $N = 2k - 1$ an odd number.  Let $w = a_{N}...... a_1$ and $r = a_1.....a_1$ be an $N$ digit number and its reversal.  We want to see if it is ever possible to have $w + r$ be such that all digits are odd.
$$w +r = (1)[a_{2k-1} + a_1(+1)\mod 10][a_{2k-2} + a_2(+1)\mod 10] ....[a_2 + a_{2k-2}(+1)\mod 10][a_1 + a_{2k-1}\mod 10]$$.
where the $(+1)$ means that maybe the term to the right added to more than $10$ and we had to "carry" a $1$; or maybe we didn't.  And the $\mod 10$ means if the term adds to more than $10$ we only take the last digit.
If all the terms $[a_{2k -i} + a_i(+1)\mod 10]$ are odd then two terms, $[a_{2k -i} + a_i(+1)\mod 10]$ and $[a_{i} + a_{2k-i} (+1)\mod 10]$ must either have both carried from the right or both have not.  And likewise either they both cause us to carry to the left or they both don't.
The middle term is $a_k$ so have $[a_k + a_k (+1)\mod 10]$.  To be odd we must have borrowed from $[a_{k-1} + a_{k+1}(+1) \mod 10]$.  Which means $[a_{k+1} + a_{k-1} (+1)\mod 10]$ must cause us to carry to $[a_{k+2} + a_{k-2}(+1)\mod 10]$.  Which in turn means $[a_{k-2} + a_{k+2}(+1)\mod 10]$ must have borrowed from $[a_{k-3} + a_{k+3}(+1)\mod 10]$.
Inductively this means every odd $i$ then $[a_{k\pm i} + a_{k\mp i} (+1)\mod 10]$ carried to the left and for every even $j$ then $[a_{k\pm j} + a_{k\mp j} (+1)\mod 10]$ borrowed from the right.
Now what about the term $[a_1 + a_{2k-1} \mod 10]= [a_{k-(k-1)} + a_{k + (k-1)} \mod 10]$, the furthest term to the right?  It can not have borrowed because it is the very first term. So $k-1$ is odd and $k$ is even.
This can only happen if $k=2m$ is even or $N = 4m - 1$ is of the form $4(m-1) + 3$.
It can not have if $k=2m +1$ is odd or $N = 4m + 2-1 = 4m + 1$.
For it to happen if $N = 4m -1$ we must have: for all odd $i$, $a_{k + i} + a_{k-i}$ be odd and greater than $10$ so it carries and for all even $j$, $a_{k+j} + a_{k-j}$ is less than $10$ and even so that when it borrows $1$ it will be odd.  This is always possible.  
Example.  Let $a_{k\pm j} = 4$ for every even $j$ and let $a_{k+i} = 5$ and $a_{k-i} = 6$ for every odd $i$ you get:  $5454....464646 + 6464646.....4545 = 1919......9191$
