6. Let $f(x)$ be defined on $\mathbf{N}_{+}$, with its range $B \subseteq$ $\mathbf{N}_{+}$, and for any $n \in \mathbf{N}_{+}$, we have
$$
f(n+1)>f(n) \text {, and } f(f(n))=3 n \text {. }
$$

Then $f(10)+f(11)=$ $\qquad$
6.39.

Given $f(f(1))=3$, we know $f(f(f(1)))=f(3)$.
If $f(1)=1$, then $3=f(f(1))=f(1)=1$, which is a contradiction.
Therefore, $2 \leqslant f(1)<f(2) \leqslant f(f(1))=3$.
Thus, $f(2)=3, f(1)=2$,
$f(3)=f(f(2))=6$,
$f(6)=f(f(3))=9$.
Also, $6=f(3)<f(4)<f(5)<f(6)=9$, so
$f(4)=7, f(5)=8$,
$f(7)=f(f(4))=12$,
$f(12)=f(f(7))=21$.
Since $f(9)=f(f(6))=18$,
$18=f(9)<f(10)<f(11)<f(12)=21$,

Therefore, $f(10)=19, f(11)=20$.
Thus, $f(10)+f(11)=39$.