6. Let the planar region $T=\{(x, y) \mid x>0, y>0, x y \leqslant 48\}$, then the number of lattice points within $T$ is $\qquad$.


$$
\begin{array}{l}
\text { 6. } 202 \text { Let } T_{1}=\left\{(x, y) \mid 0<x \leqslant \sqrt{48}, 0<y \leqslant \frac{n}{x}\right\}, \\
T_{2}=\left\{(x, y) \mid 0<y \leqslant \sqrt{48}, 0<x \leqslant \frac{n}{y}\right\} .
\end{array}
$$

Then $T=T_{1} \cup T_{2}, T_{1} \cap T_{2}=\{(x, y) \mid 0<x \leqslant \sqrt{48}, 0<y<\sqrt{48}\}$, using $\|x\|$ to denote the number of lattice points in set $x$, thus according to the principle of inclusion and exclusion, we have
$$
\begin{aligned}
\|T\| & =\left\|T_{1} \cup T_{2}\right\| \\
& =\left\|T_{1}\right\|+\left\|T_{2}\right\|-\left\|T_{1} \cap T_{2}\right\| \\
& =\sum_{0<x \leqslant \sqrt{48}}\left[\frac{48}{x}\right]+\sum_{0<y \leqslant \sqrt{48}}\left[\frac{48}{y}\right]-[\sqrt{48}]^{2} \\
& =2 \sum_{0<k \leqslant 6}\left[\frac{48}{k}\right]-36=119 \times 2-36=202 .
\end{aligned}
$$