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

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

Clearly, $k<2 n+k+1$.
To maximize the smaller factor $k$ on the right side of the equation, while ensuring $n$ is non-negative, the largest possible value is
$$
k=2 \cdot 3^{5}, 2 n+k+1=3^{6} \text {. }
$$

In this case, we can solve for $n=121$.
Thus, we have
$$
3^{11}=122+123+\cdots+607 \text {. }
$$

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