1.26 Find the maximum value of $k$ such that $3^{11}$ can be expressed as the sum of $k$ consecutive positive integers.
(5th American Invitational Mathematics Examination, 1987)
[Solution] Assume $3^{11}$ is expressed as the sum of $k$ consecutive positive integers:
$$
3^{11}=(n+1)+(n+2)+\cdots+(n+k).
$$

where $n$ is a non-negative integer, and $k$ is a positive integer.
We seek the maximum value of $k$ that satisfies (1).
$$
\begin{aligned}
3^{11} & =n k+\frac{k(k+1)}{2}, \\
2 \cdot 3^{11} & =2 n k+k(k+1), \\
2 \cdot 3^{11} & =k(2 n+k+1). \\
k & <2 n+k+1.
\end{aligned}
$$

Obviously,
to make the smaller factor $k$ on the right side of the equation as large as possible, while ensuring that $n$ is non-negative, the maximum possibility is
$$
k=2 \cdot 3^{5}, \quad 2 n+k+1=3^{6}.
$$

In this case, we can solve for
$$
n=121.
$$

Therefore, we have
$$
3^{11}=122+123+\cdots+607.
$$

The maximum $k$ sought is $2 \cdot 3^{5}=486$.