Edit: I mistook it to be the number of digits being odd or even. Thanks mszew for pointing it out

Saint123 wrote: What do you mean by a sum having 'have odd digits only'? What if $a_i=8$ for all i? The sum has even digits. Correspondingly assume $a_i=1$ then the sum has 1001 digits. Each of the digits of the sum is odd. For example for a small case with $2$ digits instead of $1001$, $12 + 21=33$ Moreover it is possible for every $2k$-digits numbers with $1212...12+2121...21=3333...33$

It is possible for every $2k$-digits numbers with $1212...12+2121...21=3333...33$ It is possible for $4k+3$-digits numbers with $516+615=1131$, $5151616+6161515=11313131$, $5151...516...1616+6161...615...1515=11313...131...3131$ It is not possible for $4k+1$-digits numbers Assume that it is possible, defining the different column sums of the numbers without the carry overs from left to right, let $b_{2k}=a_0+a_{4k}$, $b_{2k-1}=a_1+a_{4k-1}$,$...b_{1}=a_{2k-1}+a_{2k+1}$ $m_k=a_{2k}+a_{2k}$ $c_{1}=a_{2k+1}+a_{2k-1}...c_{2k-1}=a_{4k-1}+a_1, c_{2k}=a_{4k}+a_0$ First of all notice that $b_i=c_i$ Then $m_k$ is even then $c_1>9$ Step 1) $c_1>9$ then $b_1>9$ then $b_2$ is even then $c_2$ is even then $c_3>9$ Repeating that step to get $c_{2k}$ even contradiction.