F11 (29-3, UK) Let $\mathbf{N}^{*}$ be the set of positive integers, and define the function $f$ on $\mathbf{N}^{*}$ as follows:
(1) $f(1)=1, f(3)=3$;
(2) For $n \in \mathbf{N}^{*}$, we have
$$
\begin{array}{c}
f(2 n)=f(n), \\
f(4 n+1)=2 f(2 n+1)-f(n), \\
f(4 n+3)=3 f(2 n+1)-2 f(n) .
\end{array}
$$

Question: How many $n \in \mathbf{N}^{*}$, and $n \leqslant 1988$, satisfy $f(n)=n$?
If the expression of $f$ can be determined, the problem will be solved.
Let the binary representation of $n$ be $\left(a_{s} a_{s-1} \cdots a_{1}\right)_{2}$. We will prove:
$$
f(n)=\left(a_{1} a_{2} \cdots a_{s}\right)_{2}.
$$

When $n=1,2,3$, the above formula obviously holds.
Since $f(2 n)=f(n)$, we only need to consider the case when $n$ is odd. Assume (1) holds for $<n$, now consider $f(n)$.
(1) When $n=4 m+1$, let $4 m+1=\left(a_{k} a_{k-1} \cdots a_{0}\right)_{2}$, where $a_{0}=1, a_{1}=0$. Thus, $4 m=\left(a_{k} a_{k-1} \cdots a_{1} 0\right)_{2}$, which gives
$$
m=\left(a_{k} a_{k-1} \cdots a_{2}\right)_{2}, 2 m+1=\left(a_{k} a_{k-1} \cdots a_{2} 1\right)_{2}.
$$

By the induction hypothesis, we have
$$
\begin{array}{c}
f(2 m+1)=\left(1 a_{2} a_{3} \cdots a_{k}\right)_{2}, \\
f(m)=\left(a_{2} a_{3} \cdots a_{k}\right)_{2},
\end{array}
$$

Thus,
$$
\begin{aligned}
f(4 m+1) & =2 f(2 m+1)-f(m) \\
& =\left(1 a_{2} a_{3} \cdots a_{k} 0\right)_{2}-\left(a_{2} a_{3} \cdots a_{k}\right)_{2} \\
& =2^{k}+\sum_{j=2}^{k} a_{j} \cdot 2^{k+1-j}-\sum_{j=2}^{k} a_{j} \cdot 2^{k-j} \\
& =2^{k}+\sum_{j=2}^{k} a_{j} \cdot 2^{k-j} \\
& =\sum_{j=0}^{k} a_{j} \cdot 2^{k-j} \\
& =\left(a_{0} a_{1} \cdots a_{k}\right)_{2},
\end{aligned}
$$

The conclusion holds.
(2) When $n=4 m+3$, still let $4 m+3=\left(a_{k} a_{k-1} \cdots a_{0}\right)_{2}$, where $a_{0}=a_{1}=1$, thus $4 m=\left(a_{k} a_{k-1} \cdots a_{2} 00\right)_{2}, m=\left(a_{k} a_{k-1} \cdots a_{2}\right)_{2}, 2 m+1=\left(a_{k} a_{k-1} \cdots a_{2} 1\right)_{2}$.
By the induction hypothesis and the problem statement, we have
$$
\begin{aligned}
f(4 m+3) & =3 f(2 m+1)-2 f(m) \\
& =3\left(1 a_{2} \cdots a_{k}\right)_{2}-2\left(a_{2} a_{3} \cdots a_{k}\right)_{2} \\
& =\left(1 a_{2} \cdots a_{k} 0\right)_{2}+\left(1 a_{2} \cdots a_{k}\right)_{2}-\left(a_{2} a_{3} \cdots a_{k} 0\right)_{2} \\
& =2^{k}+\left(1 a_{2} \cdots a_{k}\right)_{2} \\
& =\left(11 a_{2} \cdots a_{k}\right)_{2} \\
& =\left(a_{0} a_{1} a_{2} \cdots a_{k}\right)_{2},
\end{aligned}
$$

Thus, the conclusion is proved.
Since $f(n)=n \Leftrightarrow\left(a_{1} a_{2} \cdots a_{k}\right)_{2}=\left(a_{k} a_{k-1} \cdots a_{1}\right)_{2}$, where $n=\left(a_{k} a_{k-1} \cdots a_{1}\right)_{2}$, the binary number $n$ that satisfies $f(n)=n$ is "symmetric". Therefore, if $n$ has $2 m$ binary digits, it is completely determined by the first $m$ digits. Since the first digit is 1, the following $m-1$ digits can be any of 0 or 1, so there are $2^{m-1}$ symmetric binary numbers with $2 m$ digits. Similarly, there are $2^{m-1}$ symmetric binary numbers with $2 m-1$ digits.

Since $2^{10}<1988<2^{11}$, among the positive integers less than 2048, there are
$$
1+1+2+2+4+4+8+8+16+16+32=94 \text{ (numbers). }
$$

Since $1988=(11111000100)_{2}$, the symmetric numbers exceeding 1988 are only two, they are $(11111011111)_{2}, (1111111111)_{2}$, so there are 92 positive integers $n$ not exceeding 1988 such that $f(n)=n$.