jmc

Say our two-digit non-palindrome is $n=\overline{ab}=10a+b$, with $a$ and $b$ digits. Reversing $n$ and adding it to itself is $10a+b+10b+a=11(a+b)$. This operation only depends on $a+b$, so 57 and 48 for example yield the same result. When $a+b\le 9$, the resulting number is just a number in $\{11,22,\dots ,99\}$, all of which are palindromes, so numbers with $a+b\le 9$ take one step. We can now check how many times the operation needs to be applied on each remaining value of $a+b$. Since $a,b\le 9$, $a+b\le 18$. $$a+b=10\to 110\to 121$$ $$a+b=11\to 121$$ $$a+b=12\to 132\to 363$$ $$a+b=13\to 143\to 484$$ $$a+b=14\to 154\to 605\to 1111$$ $$a+b=15\to 165\to 726\to 1353\to 4884$$ $$a+b=16\to 176\to 847\to 1595\to 7546\to 14003\to 44044$$ $$a+b=17\to 187\to 968\to 1837\to 9218\to 17347\to 91718\to \dots$$ $$a+b=18\to 198\to 1089\to 10890\to 20691\to 40293\to 79497$$ The only two values of $a+b$ which require exactly six steps are $a+b=16$ and $a+b=18$. However, the only $n$ for which $a+b=18$ is $n=99$, a palindrome. We are left with $97+79=\boxed{176}$, as we exclude the palindrome $n=88$.