6. For the geometric sequence $\left\{a_{n}\right\}$ with the first term $a_{1}=1536$, and common ratio $q=-\frac{1}{2}$, let $\pi_{n}$ denote the product of its first $n$ terms. The maximum value of $\pi_{n}$ is $\qquad$ .
6. $\pi_{12}$.

From the problem, we have $a_{n}=1536 \times\left(-\frac{1}{2}\right)^{n-1}, \pi_{n}=(1536)^{n} \times\left(-\frac{1}{2}\right)^{\frac{n(n-1)}{2}},\left|\pi_{n}\right|=(1536)^n \left(\frac{1}{2}\right)^{\frac{n(n-1)}{2}}$, and $\frac{\left|\pi_{n+1}\right|}{\left|\pi_{n}\right|}=\left|a_{n+1}\right|=1536 \cdot\left(\frac{1}{2}\right)^{n}$. When $n \leqslant 10,\left|\pi_{n+1}\right|>\left|\pi_{n}\right| ; n \geqslant 11$, $\left|\pi_{n+1}\right|<\left|\pi_{n}\right| ; \left|\pi_{12}\right|>\left|\pi_{13}\right| \cdots$. Also, $\pi_{11}>0, \pi_{10}>0, \pi_{13}>0, \pi_{12}>\pi_{13}$, so it can only be that $\pi_{12}$ and $\pi_{9}$ are the largest. And $\frac{\pi_{12}}{\pi_{9}}=\frac{27}{8}>1, \pi_{12}>\pi_{9}$, hence $\pi_{12}$ is the largest.