jmc

Let us write down one such sum, with $m$ terms and first term $n+1$: $3^{11}=(n+1)+(n+2)+\dots +(n+m)=\frac{1}{2}m(2n+m+1)$. Thus $m(2n+m+1)=2\cdot 3^{11}$ so $m$ is a divisor of $2\cdot 3^{11}$. However, because $n\ge 0$ we have $m^2<m(m+1)\le 2\cdot 3^{11}$ so $m<\sqrt{2\cdot 3^{11}}<3^6$. Thus, we are looking for large factors of $2\cdot 3^{11}$ which are less than $3^6$. The largest such factor is clearly $2\cdot 3^5=486$; for this value of $m$ we do indeed have the valid expression $3^{11}=122+123+\dots +607$, for which $k=\boxed{486}$.