abcd-dcba

Checking powers of $13$ modulo $20$

$\begin{cases} 13^1≡13\phantom{i}mod\phantom{i}20\\ 13^2≡9\phantom{i}mod\phantom{i}20\\ 13^3≡17\phantom{i}mod\phantom{i}20\\ 13^4≡1\phantom{i}mod\phantom{i}20\\ \cdots \end{cases}$

And then the pattern repeats since $13^4≡1\phantom{i}mod\phantom{i}20$

Adding up all this remainders

$13+9+17+1≡40≡0\phantom{i}mod\phantom{i}20$

Since none of the partial sums $13,13+9,13+9+17$ is divisible by $20$ , $n$ has to be a multiple of $4$ in order to have

$20\mid\sum_{k=1}^n 13^k$

Among the answers, only $\boxed{76}$ is a multiple of $4$