Example 14 (IMO-29 Problem) Let $\mathbf{N}$ be the set of positive integers, and define the function $f$ on $\mathbf{N}$ as follows:
(i) $f(1)=1, f(3)=3$;
(ii) For $n \in \mathbf{N}$, we have $f(2 n)=f(n), f(4 n+1)=2 f(2 n+1)-f(n), f(4 n+3)=$ $3 f(2 n+1)-2 f(n)$.
Find all $n$ such that $n \leqslant 1988$ and $f(n)=n$.
From conditions (i) and (ii), we can calculate the following table: converting the numbers in the table to binary, we get
Thus, we conjecture: $f(n)$ is equal to the binary number obtained by reversing the binary representation of $n$.
Next, we will prove that this conjecture is correct.
In fact, when $n=1,3$, the conjecture holds.
Assume that the conjecture holds for $n<k$. We will prove that the conjecture also holds for $n=k$.
Indeed, if $k$ is even, then we can set the binary number of $k$ as $\left(a_{l} a_{l-1} \cdots a_{1} 0\right)_{2}$, and the binary number of $\frac{k}{2}$ as $\left(a_{l} a_{l-1} \cdots a_{1}\right)_{2}$. According to the given conditions and the induction hypothesis, we have $f(k)=f\left(\frac{k}{2}\right)=\left(a_{1} \cdots a_{l-1} a_{l}\right)_{2}$. Therefore, the conjecture holds when $k$ is even.
If $k=4 m+1 (m \in \mathbf{N})$, then we can set the binary number of $k$ as
$k=4 m+1=\left(a_{l} a_{l-1} \cdots a_{2} 01\right)_{2}$,
Thus, $m=\left(a_{l} a_{l-1} \cdots a_{2}\right)_{2}$, and $2 m+1=\left(a_{l} a_{l-1} \cdots a_{2} 1\right)_{2}$,
At this point, according to the given conditions and the induction hypothesis, we have
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$$
\begin{aligned}
f(4 m+1) & =2 f(2 m+1)-f(m) \\
& =\left(1 a_{2} \cdots a_{l-1} a_{1} 0\right)_{2}-\left(a_{2} \cdots a_{l-1} a_{l}\right)_{2}=2^{l}+2 f(m)-f(m) \\
& =2^{l}+f(m)=\left(10 a_{2} \cdots a_{l-1} a_{l}\right)_{2} .
\end{aligned}
$$

This means that the conjecture also holds when $k=4 m+1$.
Similarly, it can be proven that the conjecture also holds when $k=4 m+3$.
According to the principle of mathematical induction, the conjecture is proven.
$n=\left(a_{1} a_{2} \cdots a_{l}\right)_{2}=\left(a_{l} a_{l-1} \cdots a_{1}\right)_{2}$, i.e., $a_{i}=a_{l+1-i}\left(i=1,2, \cdots,\left[\frac{l}{2}\right]\right)$.
When the number of binary digits of $n$ is even $2 s$, the number of $n$ satisfying the above condition is $2^{s-1}$; when the number of binary digits of $n$ is odd $2 s+1$, the number of $n$ satisfying the above condition is $2^{s}$.

Notice that $1988=(11111000100)_{2}$, and the 11-bit binary numbers greater than this number that satisfy $f(n)=n$ are only two: $(11111111111)_{2}$ and $(11111011111)_{2}$. Therefore, the number of such numbers is.
$$
(1+1+2+2+4+4+8+8+16+16+32)-2=92
$$