$\mathbf{F 1 7}$ (39-6, Bulgaria) Let $\mathbf{N}^{*}$ be the set of all positive integers, and the function $f: \mathbf{N}^{*} \rightarrow \mathbf{N}^{*}$ satisfies: for any $s$ and $t$ in $\mathbf{N}^{*}$, we have
$$
f\left(t^{2} f(s)\right)=s(f(t))^{2},
$$

Determine the smallest possible value of $f(1998)$ among all such functions $f$.
First, we prove the given statement:
$$
f\left(t^{2} f(s)\right)=s(f(t))^{2}
$$

is equivalent to
$$
\left\{\begin{array}{l}
f(f(s))=s(f(1))^{2}, \\
f(s t)=\frac{f(s) f(t)}{f(1)} .
\end{array}\right.
$$

From (2) and (3), we get
$$
\begin{aligned}
f\left(t^{2} f(s)\right) & =\frac{f(t) f(t f(s))}{f(1)} \\
& =\frac{f(t) \cdot f(t) f(f(s))}{(f(1))^{2}} \\
& =s(f(t))^{2},
\end{aligned}
$$

which means (1) holds.
In (1), let \( t=1 \), we get \( f(f(s))=s(f(1))^{2} \). In (1), substitute \( f(s) \) for \( s \), we get
$$
\begin{aligned}
f\left(t^{2} s(f(1))^{2}\right) & =f\left(t^{2} f(f(s))\right) \\
& =f(s)(f(t))^{2},
\end{aligned}
$$

Substitute \( s=t^{2}, t=1 \) into (4), we get
$$
\begin{aligned}
f\left(t^{2}\right)(f(1))^{2} & =f\left(t^{2}(f(1))^{2}\right) \\
& =f(1)(f(t))^{2},
\end{aligned}
$$

Thus,
$$
f\left(t^{2}\right)=\frac{(f(t))^{2}}{f(1)} \text {. }
$$

From (5), we get
$$
f\left(s^{2} t^{2}\right)=\frac{(f(s t))^{2}}{f(1)} .
$$

From (4) and (6), we get
$$
\begin{aligned}
f\left(s^{2} t^{2}\right)(f(1))^{2} & =f\left(s^{2} t^{2}(f(1))^{2}\right) \\
& =f\left(t^{2}\right)(f(s))^{2} \\
& =\frac{(f(t))^{2}}{f(1)} \cdot(f(s))^{2},
\end{aligned}
$$

From (6) and (7), we get (3).
Next, we prove: for any \( t \in \mathbf{N}^{*} \), \( f(1) \mid f(t) \).
In fact, from (3) and (5), combined with mathematical induction, we get
$$
f\left(t^{n}\right)=\left(\frac{f(t)}{f(1)}\right)^{n-1} f(t), n \in \mathbf{N}^{*} .
$$

If \( f(1) \nmid f(t) \), then there exists a prime factor \( p \) of \( f(1) \) and a positive integer \( m \), such that \( p^{m} \mid f(1) \), thus when \( n>m \), \( p^{m(n-1)} \nmid f^{n}(t) \), leading to \( f\left(t^{n}\right) \notin \mathbf{N}^{*} \), which is a contradiction.

Let \( f(m)=k_{m} \cdot f(1)\left(m \in \mathbf{N}^{*}\right) \), we can prove: when \( s \neq t \), \( k_{s} \neq k_{t} \), because if \( k_{s}=k_{1} \),
$$
s(f(1))^{2}=f(f(s))=f(f(t))=t(f(1))^{2},
$$

thus \( s=t \). In particular, when \( m>1 \), \( k_{m}>1 \).
Now, we find the minimum value of \( f(1998) \). Since
$$
\begin{aligned}
f(1998) & =f\left(2 \times 3^{3} \times 37\right) \\
& =\frac{f(2)(f(3))^{3} f(37)}{(f(1))^{4}} \\
& =k_{2} \cdot k_{3}^{3} \cdot k_{37} f(1),
\end{aligned}
$$

where \( k_{2}, k_{3}, k_{37} \) are all greater than 1 and distinct. Thus,
$$
f(1998) \geqslant 1 \times 2^{3} \times 3 \times 5=120 \text {. }
$$

In fact, we need to show that \( f(1998) \neq 1 \times 2^{3} \times 3 \times 4 \). If \( f(1998)=1 \times 2^{3} \times 3 \times 4 \), then \( f(3)=2, f(1)=1 \). At this time, \( f(9)=\frac{(f(3))^{2}}{f(1)}=4 \), so \( k_{2} \) and \( k_{37} \) cannot both be 4, which is a contradiction.

Below, we construct a function \( f: \mathbf{N}^{*} \rightarrow \mathbf{N}^{*} \) such that \( f(1998)=120 \). Let \( f(1)=1, f(2)=3, f(3)=2, f(5)=37, f(37)=5, f(p)=p \) (where \( p \) is a prime number different from 2, 3, 5, 37), and when \( n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{r}^{a_{r}} \) is a composite number, \( f(n)=f\left(p_{1}\right)^{a_{1}} f\left(p_{2}\right)^{a_{2}} \cdots f\left(p_{r}\right)^{a_{r}} \). Clearly, this function \( f \) satisfies (2) and (3).
In conclusion, the minimum value of \( f(1998) \) is 120.