2. Try to find the largest positive number $c$, such that for each natural number $n$, $\{n \sqrt{2}\} \geqslant \frac{c}{n}$, and determine the natural number $n$ for which $\{n \sqrt{2}\}=\frac{c}{n}$.
2. Solution: The required maximum positive number $c=\frac{\sqrt{2}}{4}$.

If $c>0$, then for $n \in \mathbf{N}$, we have $\{n \sqrt{2}\} \geqslant \frac{c}{n}$.
Let $n_{k}=\frac{(\sqrt{2}+1)^{4 k+1}+(\sqrt{2}-1)^{4 k+1}}{2 \sqrt{2}}$, then $n_{k} \in \mathbf{N}_{+}$, 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 $m > n \sqrt{2} - 1$.
Thus, $1 \leqslant 2 n^{2} - m^{2} = n^{2}\left(\sqrt{2} + \frac{m}{n}\right)\left(\sqrt{2} - \frac{m}{n}\right)2 \sqrt{2}$.
Therefore, for all $n \in \mathbf{N}$,
$$
\frac{1}{2 \sqrt{2} n} < n\left(\sqrt{2} - \frac{m}{n}\right) = n \sqrt{2} - m = \{n \sqrt{2}\} \text{. }
$$

In summary, $c_{\max} = \frac{\sqrt{2}}{4}$.