7. (10 points) If $2 \times 3^{8}$ can be expressed as the sum of $k$ consecutive positive integers, then the maximum value of $k$ is

Express the above text in English, keeping the original text's line breaks and format, and output the translation result directly.
【Analysis】First, we can set out $k$ consecutive positive integers, find their sum, and focus on maximizing $k$ for the solution.
【Solution】Solution: Let the $k$ consecutive integers be $n+1, n+2, n+3, \cdots, n+k$, then the sum $=\frac{\mathrm{k} \times(\mathrm{n}+1+\mathrm{n}+\mathrm{k})}{2}=\frac{\mathrm{k} \times(2 \mathrm{n}+\mathrm{k}+1)}{2}$. Since $k$ is maximized, $n$ is minimized, and $k<2 n+k+1$, $\frac{k \times(2 n+k+1)}{2}=2 \times 3^{8}$, which means $k \times(2 n+k+1)=2^{2} \times 3^{8}=\left(2^{2} \times 3^{4}\right) \times 3^{4}=3^{5} \times\left(2^{2} \times 3^{3}\right)$, thus the maximum value of $k$ is $3^{4}=108$. Therefore, the answer is: 108.