6・136 Let $N$ be the set of natural numbers. Let $f: N \rightarrow N$ satisfy the conditions: $f(1)=1$, and for any natural number $n$,
$$
\left\{\begin{array}{l}
3 f(n) f(2 n+1)=f(2 n)(1+3 f(n)), \\
f(2 n)<6 f(n) .
\end{array}\right.
$$

Find all solutions to the equation
$$
f(k)+f(l)=293, k<l
$$
[Solution] From the given conditions, for all $n \in N$, we have
$$
1 \leqslant 3 f(n)(f(2 n+1)-f(2 n))=f(2 n)<6 f(n) .
$$

Therefore,
$$
\begin{array}{l}
1 \leqslant f(2 n+1)-f(2 n)<2 \\
f(2 n+1)=f(2 n)+1
\end{array}
$$

Substituting (2) into (1) yields
$$
f(2 n)=3 f(n) .
$$

Using (2) and (3), we can prove by mathematical induction that
$$
f\left(2^{n_{1}}+2^{n_{2}}+\cdots+2^{n_{s}}\right)=3^{n_{1}}+3^{n_{2}}+\cdots+3^{n_{s}},
$$

where the non-negative integers $n_{1}, n_{2}, \cdots, n_{s}$ satisfy
$$
0 \leqslant n_{1}<n_{2}<\cdots<n_{s} \text {. }
$$

Let
$$
\begin{array}{l}
k=2^{n_{1}}+2^{n_{2}}+\cdots+2^{n_{s}}, \\
l=2^{m_{1}}+2^{m_{2}}+\cdots+2^{m_{t}},
\end{array}
$$

where $n_{1}<n_{2}<\cdots<n_{s}, m_{1}<m_{2}<\cdots<m_{t}$.
When $k<l$, it is clear that $f(k)<f(l)$. Note that
$$
293=3^{5}+3^{3}+2 \times 3^{2}+3+2 \times 3^{0} \text {, }
$$

where $3^{5}=243$ must be one of the addends of $f(l)$, and $3^{2}$ and $3^{0}$ must be addends in both $f(k)$ and $f(l)$, while $3^{3}$ and 3 are uncertain. Therefore, $f(k)$ and $f(l)$ can only have the following 4 cases:
$$
\begin{array}{l}
\left\{\begin{array}{l}
f(k)=3^{2}+3^{0} \\
f(l)=3^{5}+3^{3}+3^{2}+3+3^{0}
\end{array}\right. \\
\left\{\begin{array}{l}
f(k)=3^{3}+3^{2}+3^{0} \\
f(l)=3^{5}+3^{2}+3+3^{0}
\end{array}\right. \\
\left\{\begin{array}{l}
f(k)=3^{2}+3+3^{0} \\
f(l)=3^{5}+3^{3}+3^{2}+3^{0}
\end{array}\right. \\
\left\{\begin{array}{l}
f(k)=3^{3}+3^{2}+3+3^{0} \\
f(l)=3^{5}+3^{2}+3^{0}
\end{array}\right.
\end{array}
$$

Accordingly, we obtain all solutions to the given equation:
$$
\begin{array}{l}
\left\{\begin{array}{l}
k=2^{2}+2^{0}=5, \\
l=2^{5}+2^{3}+2^{2}+2+1=47 ;
\end{array}\right. \\
\left\{\begin{array}{l}
k=2^{3}+2^{2}+2^{0}=13, \\
l=2^{5}+2^{2}+2+2^{0}=39 ;
\end{array}\right. \\
\left\{\begin{array}{l}
k=2^{2}+2+2^{0}=7, \\
l=2^{5}+2^{3}+2^{2}+2^{0}=45 ;
\end{array}\right. \\
\left\{\begin{array}{l}
k=2^{3}+2^{2}+2+2^{0}=15, \\
l=2^{5}+2^{2}+2^{0}=37 .
\end{array}\right.
\end{array}
$$