II. (15 points) The function $f(k)$ is defined on $N$ and takes values in $N$, being 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 $f(f(k))=3k$. Try to find the value of $f(1)+f(9)+f(96)$.
For $k \in N, f(f(k))=3_{i}^{2}$.
(1)
$\therefore f[f(f(\hat{i})]:=\hat{j}(3 k)$
Also, $f[f(f(i))]=3 f(k)$,
$\therefore f(3 k)=3 f(k)$.
If $f(1)=1$, substituting into (1) gives $f(1)=3$, a contradiction.
$\therefore f(1)=a>1$, but $f(f(1))=f(a)=3$.
By the strict monotonicity of $f(k)$, i.e., $1<a \Rightarrow f(1)<f(a)=3$,
$\therefore$ it can only be that $f(1)=2, f(2)=3$.
At this point, $f(3)=3 f(1)=6$,
$$
\begin{array}{l}
f(6)=f(3 \times 2)=3 f(2)=9, \\
f(9)=3 f(3)=18, \\
f(18)=3 f(6)=27, \\
f(27)=3 f(9)=54, \\
f(54)=3 f(18)=81 .
\end{array}
$$

Note, as the variable changes from 27 to 54, it increases by 27 numbers, and the function value correspondingly increases from 54 to 81, also by 27 numbers. By the strict monotonicity of $f$,
$$
\begin{array}{l}
f(28)=55, f(29)=56, f(30)=57, \\
f(31)=58, f(32)=59, \cdots .
\end{array}
$$

And $f(96)=f(3 \times 32)=3 f(32)=3 \times 59=177$.
$$
\therefore f(1)+f(9)+f(96)=2+18+177=197 \text {. }
$$