(1) Try to find the largest positive integer $k$, such that there exists a triangle with side lengths all being positive integers not greater than $n$, and the difference between any two sides (larger minus smaller) is not less than $k$;
(2) Try to find all positive integers $n(n \geqslant 4)$, such that the triangle corresponding to the largest positive integer $k$ in (1) is unique.
(1) Let the lengths of the three sides of a triangle be positive integers $a, b, c$, and
$1 \leqslant a < b < c \leqslant n, a + b \geqslant c + 1$,
$c - b \geqslant k, b - a \geqslant k$.
Then, $b \leqslant c - k, a \leqslant b - k \leqslant c - 2k$.
Thus, $(c - 2k) + (c - k) \geqslant a + b \geqslant c + 1$.
Hence, $3k + 1 \leqslant c \leqslant n$.
Therefore, $k \leqslant \frac{n-1}{3}$.
Since $k \in \mathbf{N}$, thus, $k \leqslant \left[\frac{n-1}{3}\right]$.
When $k = \left[\frac{n-1}{3}\right]$, we can take a positive integer $c$ such that $3 \times \left[\frac{n-1}{3}\right] + 1 \leqslant c \leqslant n$, and then take $b = c - \left[\frac{n-1}{3}\right], a = c - 2 \times \left[\frac{n-1}{3}\right]$, at this time $a, b, c \in \mathbf{N}$, and satisfy $c - b \geqslant k, b - a \geqslant k, a + b \geqslant c + 1$. This shows that when $k = \left[\frac{n-1}{3}\right]$, there exists a triangle that meets the requirements.
In summary, the maximum value of the positive integer $k$ is $\left[\frac{n-1}{3}\right]$.
(2) From (1), we know $3k + 1 \leqslant c \leqslant n$, where $k = \left[\frac{n-1}{3}\right]$, so,
$$
3 \times \left[\frac{n-1}{3}\right] + 1 \leqslant c \leqslant n.
$$

To make the triangle corresponding to $k = \left[\frac{n-1}{3}\right]$ unique, it must be that $3 \times \left[\frac{n-1}{3}\right] + 1 = n$ (otherwise, $c$ can take 2 positive integers), i.e., $\left[\frac{n-1}{3}\right] = \frac{n-1}{3}$. Therefore, 3 divides $(n-1)$, i.e., $n \equiv 1 \pmod{3}$ and $n \geqslant 4$. Of course, this is necessary.
Below, we prove that this is also sufficient.
When $n \equiv 1 \pmod{3}$, 3 divides $(n-1)$, $c = n$, $a + b \geqslant n + 1$, $n - b \geqslant \left[\frac{n-1}{3}\right]$, $b - a \geqslant \left[\frac{n-1}{3}\right]$, $1 \leqslant a < b < c = n$, so,
$$
\begin{array}{l}
n - \left[\frac{n-1}{3}\right] \geqslant b, \\
a \leqslant b - \left[\frac{n-1}{3}\right] \leqslant n - 2 \times \left[\frac{n-1}{3}\right], \\
a + b \leqslant 2n - 3 \times \left[\frac{n-1}{3}\right] = 2n - (n-1) = n + 1.
\end{array}
$$

Also, $a + b \geqslant n + 1$, so, $a + b = n + 1$, and all the above inequalities must hold with equality.
Thus, $a = n - 2 \times \left[\frac{n-1}{3}\right] = n - \frac{2}{3}(n-1) = \frac{n+2}{3}$, $b = n - \left[\frac{n-1}{3}\right] = n - \frac{n-1}{3} = \frac{2n+1}{3}$, $c = n$.
There is only one triangle that meets the requirements, and the sufficiency is proven.
In conclusion, the required $n$ must be a positive integer satisfying $n \equiv 1 \pmod{3}$ and $n \geqslant 4$.
Note: $[x]$ represents the greatest integer not greater than $x$.