8. Let $n$ be a positive integer, and $n \geqslant 4$. 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 $k$.
8. 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 we have $b \leqslant c-k, a \leqslant b-k \leqslant c-2 k$. Therefore, $(c-2 k)+(c-k) \geqslant a+b \geqslant c+1$. Hence $3 k+1 \leqslant c \leqslant n$. So $k \leqslant \frac{n-1}{3}$. Since $k \in \mathbf{N}$, therefore, $k \leqslant\left[\frac{n-1}{3}\right]$. 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 $\left[\frac{n-1}{3}\right]$.