Answer: 486

Small Hint:

Write the sum as $\frac{k(2a+k-1)}{2}$

Big Hint:

The only possible lengths divide $2\cdot3^{11}$; then enforce a positive first term

Solution:

If the first term is $a,$ then $2\cdot3^{11}=k(2a+k-1).$ Thus $k$ is $3^j$ or $2\cdot3^j.$ The largest viable even choice is $k=2\cdot3^5=486,$ for which $2a+k-1=3^6=729$ and $a=122>0.$ The next candidates, $3^6$ and $2\cdot3^6,$ force a nonpositive first term, as do all larger choices. Hence $k=486.$