A function in itself, and for any $s$ and $t$ in $N$, it satisfies $f\left(t^{2} f(s)\right)=s(f(t))^{2}$. Determine the minimum value that $f(1998)$ can achieve among all functions $f$.

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Solution: For every function $f$ that satisfies the conditions of the problem,
it is known that $f\left(t^{2} f(s)\right)=s(f(t))^{2}$.
Substituting $t=1$ yields $f(f(s))=s(f(1))^{2}$,
and substituting $s=1$ into (1) gives
$f\left(t^{2} f(1)\right)=(f(t))^{2}$.
Now, let $f(1)=k$, then the above two equations can be rewritten as
$f(f(s))=k^{2} s$,
$f\left(k t^{2}\right)=(f(t))^{2}$.
Substituting $s=1$ into (4) gives
$f(k)=k^{2}$.
Substituting $s=f(k)$ into (1) gives
$f\left(t^{2} k^{3}\right)=k^{2}(f(t))^{2}$.
On the other hand, from (5)
$f\left(t^{2} k^{3}\right)=f\left(k(k t)^{2}\right)=(f(k t))^{2}$.
Comparing the two equations, we get
$$
(f(k t))^{2}=k^{2}(f(t))^{2} \Rightarrow f(k t)=k f(t) \text {. }
$$

Substituting (6) into (5) gives
$k f\left(t^{2}\right)=(f(t))^{2}$.
Similarly, we can write
$k f\left(t^{4}\right)=\left(f\left(t^{2}\right)\right)^{2}$,
$k f\left(t^{2^{m}}\right)=\left(f\left(t^{2^{m-1}}\right)\right)^{2}$.
Squaring the first equation and comparing it with the second gives $k^{3} f\left(t^{4}\right)=(f(t))^{4}$.
Squaring again and comparing with the third equation gives
$k^{7} f\left(t^{8}\right)=(f(t))^{8}$.
Continuing this process, we finally get
$$
(f(t))^{2^{m}}=k^{2^{m}-1} f\left(t^{2^{m}}\right) \text {. }
$$

Here $m$ is any positive integer.
Taking any prime factor $p$ of $k$, we have $p^{a} \| k$, and $p^{g} \| f(t)$.
From (7), we get $\beta \cdot 2^{m} \geqslant \alpha\left(2^{m}-1\right)$,

which implies $\frac{\beta}{\alpha} \geqslant \frac{2^{m}-1}{2^{m}}$.
Taking the limit on both sides gives
$$
\frac{\beta}{\alpha} \geqslant \lim _{m \rightarrow \infty} \frac{2^{m}-1}{2^{m}}=1 \Rightarrow \beta \geqslant \alpha \text {. }
$$

This, along with the arbitrariness of $p$, implies $k \mid f(t)$.
Now, let $g(t)=\frac{f(t)}{k}$. Then $g(t)$ is also a function from $N \rightarrow N$.
From (6), we get
$f\left(i^{2} \hat{j}(s)\right)=f\left(i_{i}^{2} \cdot k g(s)\right)=k f\left(t^{2} \cdot g(s)\right)$
$$
=b^{2} \cdot g\left(t^{2} \cdot g(s)\right)
$$

and $s(f(t))^{2}=k^{2} s(g(t))^{2}$.
From (1), we get
$$
g\left(t^{2} \cdot g(s)\right)=s(g(t))^{2} .
$$

This shows that $g$ is also a function that satisfies the conditions of the problem, and $g(1)=1$. Thus, all the results obtained for $f$ apply to $g$, with $k$ replaced by 1. Therefore, from (4) and (5), we get
$$
\begin{array}{l}
g(g(s))=s, \\
g\left(t^{2}\right)=(g(t))^{2} \text {. } \\
\text { Substituting } g(s) \text { for } s \text { in }(8) \text { gives } \\
g\left(t^{2} s\right)=g(s) \cdot(g(t))^{2} . \\
\text { Applying }(10) \text { and }(11) \text { gives } \\
(g(a b))^{2}=g\left(a^{2} b^{2}\right)=g\left(a^{2}\right)(g(b))^{2} \\
=(g(a) \cdot g(b))^{2} \\
\Rightarrow g(a b)=g(a) g(b) .
\end{array}
$$

From (9), we see that $g$ is injective, so for all $a>1$, we have $g(a)>1$. Furthermore, from (11), the function values of composite numbers are also composite, so for prime $p$, $g(g(p))=p \Rightarrow g(p)$ is prime.
Thus, we have
$$
\begin{array}{l}
f(1998)=k g(1998)=k g(2) \cdot g(3)^{3} \cdot g(37) \\
\geqslant\left(2^{3} \cdot 3 \cdot 5\right) k=120 k \geqslant 120 .
\end{array}
$$

On the other hand, consider a function $f$ defined as follows:
$$
\left\{\begin{array}{l}
f(1)=1, \\
f(2)=3, \\
f(3)=2, \\
f(5)=37, \\
f(37)=5, \\
f(p)=p(p \text { is any other prime }), \\
f(n)=f\left(p_{1}\right)^{\alpha_{1}} f\left(p_{2}\right)^{\alpha_{2} \cdots} f\left(p_{k}\right)^{a_{k}} \\
\left(n=p_{1}^{\alpha_{1}} \cdots p_{k}^{\alpha_{k}} \text { is the prime factorization of the composite number } n\right. \text {) }
\end{array}\right.
$$

It is easy to verify that it satisfies the conditions of the problem, and $f(1998)=120$. Therefore, the minimum value of $f(1998)$ is indeed 120.