$14 \cdot 7$ Try to find the largest positive number $c$, such that for every natural number $n$, $\{n \sqrt{2}\}$ $\geqslant \frac{c}{n}$. Determine the natural number $n$ for which $\{n \sqrt{2}\}=\frac{c}{n}$.
(Here $\{n \sqrt{2}\}=n \sqrt{2}-[n \sqrt{2}], \quad[x]$ denotes the integer part of $x$)
(30th International Mathematical Olympiad Candidate Problem, 1989)
[Solution] We prove that the maximum positive number $c=\frac{1}{2 \sqrt{2}}$.
If $c$ is a positive number, for $n \in \mathbb{N}$, we always have
$$
\{n \sqrt{2}\} \geqslant \frac{c}{n} \text {. }
$$

We take
$$
n_{k}=\frac{(\sqrt{2}+1)^{4 k+1}+(\sqrt{2}-1)^{4 k+1}}{2 \sqrt{2}},
$$

then $n_{k}$ is a natural number, and
$$
\begin{aligned}
\sqrt{2} n_{k}= & \frac{1}{2}\left((\sqrt{2}+1)^{4 k+1}+(\sqrt{2}-1)^{4 k+1}\right) \\
& =\frac{1}{2}\left((\sqrt{2}+1)^{4 k+1}-(\sqrt{2}-1)^{4 k+1}\right)+(\sqrt{2}-1)^{4 k+1} .
\end{aligned}
$$

Since $0 < (\sqrt{2}-1)^{4 k+1} < 1$, we have
$$
\begin{aligned}
1 & \leqslant 2 n^{2}-m^{2} \\
& =n^{2}\left(\sqrt{2}+\frac{m}{n}\right)\left(\sqrt{2}-\frac{m}{n}\right) \\
& <n^{2}\left(\sqrt{2}-\frac{m}{n}\right) \cdot 2 \sqrt{2}
\end{aligned}
$$

Thus, for all $n$,
$$
\frac{1}{2 \sqrt{2} n}<n\left(\sqrt{2}-\frac{m}{n}\right)=n \sqrt{2}-m=\{n \sqrt{2}\} \text {. }
$$

Combining the above discussions, we know that
$$
c=\frac{1}{2 \sqrt{2}} \text {. }
$$

And there does not exist an $n$ such that $\{n \sqrt{2}\}=\frac{1}{2 \sqrt{2} n}$.