6. If the difference between two positive integers is 21, and the sum of their greatest common divisor and least common multiple is 1463, then these two positive integers are $\qquad$
6. 112, 91.

Let these two positive integers be $x, y(x>y)$, their greatest common divisor be $d$, and $x=d x_{1}, y=d y_{1}$. Then their least common multiple is $d x_{1} y_{1}$, and
$$
\left\{\begin{array}{l}
d\left(x_{1}-y_{1}\right)=21, \\
d\left(1+x_{1} y_{1}\right)=1463 .
\end{array}\right.
$$

Since $(1463,21)=7$, therefore, $d=1$ or 7.

Thus $x_{1}+y_{1}=\sqrt{6289}$, which is a contradiction.
If $d=7$, then $\left\{\begin{array}{l}x_{1}-y_{1}=3, \\ x_{1} y_{1}=208 \text {. }\end{array}\right.$
Hence $x_{1}+y_{1}=\sqrt{841}=29$.
Therefore, $\left\{\begin{array}{l}x_{1}=16, \\ y_{1}=13\end{array} \Rightarrow\left\{\begin{array}{l}x=112, \\ y=91 .\end{array}\right.\right.$
So, these two positive integers are 112, 91.