11.19 Let \( N \) be the set of positive integers, and define the function \( f \) on \( N \) as follows:
(i) \( f(1)=1, \quad f(3)=3 \);
(ii) For \( n \in N \), we have
\[
\begin{array}{l}
f(2 n)=f(n), \\
f(4 n+1)=2 f(2 n+1)-f(n), \\
f(4 n+3)=3 f(2 n+1)-2 f(n) .
\end{array}
\]

Find all \( n \) such that \( n \leqslant 1988 \) and \( f(n)=n \).
(29th International Mathematical Olympiad, 1988)
\[ \text{Solution} \] From the given formula, we can obtain the following table:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline $n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\
\hline $f(n)$ & 1 & 1 & 3 & 1 & 5 & 3 & 7 & 1 & 9 & 5 & 13 & 3 & 11 & 7 & 15 & 1 & 17 \\
\hline
\end{tabular}

This table shows that, for $n \leqslant 17$, we have
\[
f\left(2^{k}\right)=1, \quad f\left(2^{k}-1\right)=2^{k}-1, \quad f\left(2^{k}+1\right)=2^{k}+1.
\]

This inspires us to consider the problem from a binary perspective, rewriting the table in binary form.
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}
\hline $n$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
\hline $f(n)$ & 1 & 1 & 3 & 1 & 5 & 3 & 7 & 1 & 9 \\
\hline $n_{(2)}$ & 1 & 10 & 11 & 100 & 101 & 110 & 111 & 1000 & 1001 \\
\hline $f(n)_{(2)}$ & 1 & 01 & 11 & 001 & 101 & 011 & 111 & 0001 & 1001 \\
\hline
\end{tabular}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline $n$ & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\
\hline $f(n)$ & 5 & 13 & 3 & 11 & 7 & 15 & 1 & 17 \\
\hline $n_{(2)}$ & 1010 & 1011 & 1100 & 1101 & 1110 & 1111 & 10000 & 10001 \\
\hline $f(n)_{(2)}$ & 0101 & 1101 & 0011 & 1011 & 0111 & 1111 & 00001 & 10001 \\
\hline
\end{tabular}

From this, we conjecture that the value of $f(n)$ is equal to the binary number formed by the reverse arrangement of the binary expansion of $n$.
We will prove this conjecture using mathematical induction.
Since $f(2n) = f(n)$, we only need to consider the case where $n$ is odd.
(1) If $n$ is of the form $4m + 1$.
Let $4m + 1 = a_k \cdot 2^k + a_{k-1} \cdot 2^{k-1} + \cdots + a_1 \cdot 2 + a_0 \cdot 1$,

where $a_i \in \{0, 1\}$, $i = 0, 1, 2, \cdots, k$, and $a_k = 1$.
Clearly, $a_0 = 1$, $a_1 = 0$.
So,
\[
\begin{array}{c}
4m = a_k \cdot 2^k + a_{k-1} \cdot 2^{k-1} + \cdots + a_2 \cdot 2^2, \\
m = a_k \cdot 2^{k-2} + a_{k-1} \cdot 2^{k-3} + \cdots + a_3 \cdot 2 + a_2, \\
2m + 1 = a_k \cdot 2^{k-1} + a_{k-1} \cdot 2^{k-2} + \cdots + a_3 \cdot 2^2 + a_2 \cdot 2 + 1.
\end{array}
\]

By the induction hypothesis,
\[
\begin{array}{l}
f(m) = a_k + a_{k-1} \cdot 2 + \cdots + a_3 \cdot 2^{k-3} + a_2 \cdot 2^{k-2}, \\
f(2m + 1) = a_k + a_{k-1} \cdot 2 + \cdots + a_3 \cdot 2^{k-3} + a_2 \cdot 2^{k-2} + 2^{k-1}.
\end{array}
\]

Now, we calculate $f(4m + 1)$.
\[
\begin{aligned}
& f(4m + 1) \\
= & 2 f(2m + 1) - f(m) \\
= & \left(a_k \cdot 2 + a_{k-1} \cdot 2^2 + \cdots + a_3 \cdot 2^{k-2} + a_2 \cdot 2^{k-1} + 2^k\right) - \left(a_k + a_{k-1} \cdot 2 + \cdots + a_3 \cdot 2^{k-3} + a_2 \cdot 2^{k-2}\right) \\
= & a_k + a_{k-1} \cdot 2 + \cdots + a_3 \cdot 2^{k-3} + a_2 \cdot 2^{k-2} + a_1 \cdot 2^{k-1} + a_0 \cdot 2^k.
\end{aligned}
\]

Thus, $f(4m + 1)$ is exactly the reverse arrangement of the binary expansion of $4m + 1$.
Therefore, the conjecture holds for $n = 4m + 1$.
(2) If $n$ is of the form $4m + 3$.

Let $4m + 3 = a_k \cdot 2^k + a_{k-1} \cdot 2^{k-1} + \cdots + a_3 \cdot 2^3 + a_2 \cdot 2^2 + a_1 \cdot 2 + a_0$,

where $a_i \in \{0, 1\}$, and $a_k = 1$. Clearly, $a_0 = a_1 = 1$.
So, $4m = a_k \cdot 2^k + a_{k-1} \cdot 2^{k-1} + \cdots + a_2 \cdot 2^2$,
\[
\begin{array}{c}
m = a_k \cdot 2^{k-2} + a_{k-1} \cdot 2^{k-3} + \cdots + a_2, \\
2m + 1 = a_k \cdot 2^{k-1} + a_{k-1} \cdot 2^{k-2} + \cdots + a_2 \cdot 2 + 1.
\end{array}
\]

By the induction hypothesis,
\[
\begin{array}{l}
f(m) = a_k + a_{k-1} \cdot 2 + \cdots + a_2 \cdot 2^{k-2}, \\
f(2m + 1) = a_k + a_{k-1} \cdot 2 + \cdots + a_2 \cdot 2^{k-2} + 2^{k-1}.
\end{array}
\]

Now, we calculate $f(4m + 3)$.
\[
\begin{aligned}
& f(4m + 3) \\
= & 3 f(2m + 1) - 2 f(m) \\
= & a_k + a_{k-1} \cdot 2 + \cdots + a_2 \cdot 2^{k-2} + 3 \cdot 2^{k-1} \\
= & a_k + a_{k-1} \cdot 2 + \cdots + a_2 \cdot 2^{k-2} + 2^{k-1} + 2^k \\
= & a_k + a_{k-1} \cdot 2 + \cdots + a_2 \cdot 2^{k-2} + a_1 \cdot 2^{k-1} + a_0 \cdot 2^k.
\end{aligned}
\]

Thus, $f(4m + 3)$ is exactly the reverse arrangement of the binary expansion of $4m + 3$.
Therefore, the conjecture holds for $n = 4m + 3$.
From the above, the conjecture holds for all $n \in \mathbb{N}$.
Let $n = (\overline{t_1 t_2 \cdots t_m})_2$ be the binary representation of $n$, then the number $n$ satisfying $f(n) = n$ should have the following symmetric form:
\[
(\overline{t_1 t_2 \cdots t_m})_2 = (\overline{t_m \cdots t_2 t_1})_2.
\]

It is easy to verify that, whether $m = 2m_1$ or $m = 2m_1 - 1$, the number of $m$-bit binary symmetric numbers is exactly $2^{m_1 - 1}$.
Therefore, the number of binary symmetric numbers less than $2^{11} = 2048$ is
\[
2(2^0 + 2^1 + 2^2 + 2^3 + 2^4) + 2^5 = 94 \text{ (numbers).}
\]

Since $1988 = (11111000100)_2$, and the binary symmetric numbers satisfying $1988 < n < 2048$ are only $(11111011111)_2$ and $(1111111111)_2$, the number of $n \leqslant 1988$ satisfying $f(n) = n$ is 92.