6. 27 The function $f(k)$ is defined on $N$, taking values in $N$, and 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 satisfies the condition $f(f(k))=3 k$. Try to find: $f(1)+f(9)$ $+f(96)$.

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

Also, by the given condition $f(f(f(k))=3 f(k)$,
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 have
$$3=f(a)>f(1)>1$$

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

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