6-19 The set of all positive integers can be divided into two disjoint subsets of positive integers $\{f(1), f(2), \cdots, f(n), \cdots\},\{g(1), g(2), \cdots, g(n), \cdots\}$,
where $\quad f(1)<f(2)<\cdots<f(n)<\cdots$
$$g(1)<g(2)<\cdots<g(n)<\cdots$$

and $\quad g(n)=f(f(n))+1 \quad(n \geqslant 1)$
Find: $f(240)$.
[Solution] First, let's prove the following three equations:
$$\begin{array}{l}
g(n)=f(n)+n \\
f(f(n))=f(n)+n-1 \\
f(f(n)+1)=f(n)+n+1
\end{array}$$

In fact, assume $f(n)=k$, then $g(n)=f(k)+1$. Thus, the two disjoint sets
$$\{f(1), f(2), f(3), \cdots, f(k)\} \text { and }\{g(1), g(2), g(3), \cdots, g(n)\}$$

contain all natural numbers from 1 to $g(n)$. By counting the number of elements in these two sets, we get

i.e., $\square$
$$g(n)=k+n$$
$$g(n)=f(n)+n$$

Thus, equation (1) holds.
Also, from $k+n=g(n)=f(k)+1$, we get
$$f(k)=k+n-1,$$

i.e.,
$$f(f(n))=f(n)+n-1$$

Thus, equation (2) holds.
$$\begin{array}{l}
\text { Let } F=\{f(n)\}, G=\{g(n)\}, n=1,2,3, \cdots . \text { From the equation } \\
\quad g(n)-1=f(f(n))
\end{array}$$

we know that $g(n)-1$ is an element of $F$. Therefore, it is impossible for two consecutive integers to both be elements of $G$.

Since $k+n$ is an element of $G$, $k+n-1$ and $k+n+1$ are both elements of $F$ and are two consecutive elements of $F$. From (2), we have

i.e.,
$$f(f(n)+1)=k+n+1$$

Thus, equation (3) holds.
Next, we use equations (1), (2), and (3) to find $f(240)$ step by step.
In fact, from $g(1)=f(f(1))+1>1$ we get $f(1)=1$, thus $g(1)=2$.
From (3), we get
$$\begin{array}{l}
f(2)=f(f(1)+1)=f(1)+1+1=3 \\
f(3)=f(f(2))=f(2)+2-1=4 \\
f(4)=f(f(3))=f(3)+3-1=6 \\
f(6)=f(f(4))=f(4)+4-1=9 \\
f(9)=f(f(6))=f(6)+6-1=14 \\
f(14)=f(f(9))=f(9)+9-1=22 \\
f(22)=f(f(14))=f(14)+14-1=35 \\
f(35)=f(f(22))=f(22)+22-1=56 \\
f(56)=f(f(35))=f(35)+35-1=90
\end{array}$$

From (2), we get

From (3), we get $\square$
$$\begin{array}{l}
f(91)=f(f(56)+1)=f(56)+56+1=147 \\
f(148)=f(f(91)+1)=f(91)+91+1=239 \\
f(240)=f(f(148)+1)=f(148)+148+1=388
\end{array}$$