6-27 The function $f(k)$ is defined on $N$ and takes values in $N$, and it is a strictly increasing function (if for any $x_{1}, x_{2} \in A$, when $x_{1}<x_{2}$, we have $f\left(x_{1}\right)<f\left(x_{2}\right)$, then $f(x)$ is called a strictly increasing function on $A$), and it satisfies the condition $f(f(k))=3k$. Try to find: $f(1)+f(9)+f(96)$.

［Solution］For any $k \in N$, by the given information,
$$
\begin{array}{l}
f(f(k))=3 k, \\
f(f(f(k)))=f(3 k) .
\end{array}
$$

Also, by the given $f(f(f(k))=3 f(k)$,
so $f(3 k)=3 f(k)$.
If $f(1)=1$, then substituting into (1) gives $f(1)=3$, which is a contradiction.

Therefore, $f(1)=a>1$.
But from (1), $f(f(1))=f(a)=3$
By the strict monotonicity of $f(x)$, we get
$$
3=f(a)>f(1)>1 \text {, }
$$

So $f(1)=2, a=2, f(2)=3$.
From (2), we get $f(3)=3 f(1)=6$,
$$
\begin{array}{l}
f(6)=3 f(2)=9, \\
f(9)=3 f(3)=18,
\end{array}
$$
$$
\begin{array}{l}
f(18)=3 f(6)=27, \\
f(27)=3 f(9)=54, \\
f(54)=3 f(18)=81 .
\end{array}
$$

From (3) and (4), as the independent variable increases from 27 to 54, by 27 numbers, the corresponding function values increase from 54 to 81, also by 27 numbers. By the strict monotonicity of the function $f(x)$, it is known that when the independent variable increases by 1 from 27 to 54, the corresponding function value also increases by 1. Therefore,
$$
f(28)=55, f(29)=56, f(30)=57, f(31)=58, f(32)=59 .
$$

From (2), we get
$$
f(96)=3 f(32)=177 .
$$

Therefore,
$$
f(1)+f(9)+f(96)=2+18+177=197 .
$$