Three, (50 points) Given positive integers $k, a, b, (a, b)=1$, find the maximum and minimum integer values of $n$ such that the linear Diophantine equation $a x + b y = n$ has exactly $k$ distinct positive integer solutions $(x, y)$.
Three, Solution: Since $(a, b)=1$, there exist $p, q \in \mathbf{Z}$ such that $p a+q b = 1$. Therefore, the equation $a x+b y=n$ has the general solution
$$
(x, y)=(p n-b t, q n+a t)(t \in \mathbf{Z}).
$$

Let $x>0, y>0$, it is easy to see that $-\frac{q n}{a}<t<\frac{p n}{b}$. Hence, the equation $a x + b y = n$ has exactly $k$ different positive integer solutions if and only if the interval $\left(-\frac{q n}{a}, \frac{p n}{b}\right)$ contains $k$ consecutive positive integers $t_{1}, t_{2}, \cdots, t_{k}$ satisfying $t_{1}-1 \leqslant-\frac{q n}{a}<t_{1}<t_{2}<\cdots<t_{k}<\frac{p n}{b} \leqslant t_{k}+1$.

Thus, the length of the interval $S=\frac{p n}{b}-\left(-\frac{q n}{a}\right)=\frac{n}{a b}$ satisfies $S \leqslant\left(t_{k}+1\right)-\left(t_{1}-1\right)=k+1$.
$$
\begin{array}{l} 
\text { Also, } S=\left(t_{k}-t_{1}\right)+\left(t_{1}-\left(-\frac{q n}{a}\right)\right)+ \\
\left(\frac{p n}{b}-t_{k}\right)=(k-1)+\frac{a t_{1}+q n}{a}+\frac{p n-b t_{k}}{b} \geqslant k-1+ \\
\frac{1}{a}+\frac{1}{b} .
\end{array}
$$

Therefore, we have $k-1+\frac{1}{a}+\frac{1}{b} \leqslant \frac{n}{a b} \leqslant k+1 \Leftrightarrow (k-1) a b + a + b \leqslant n \leqslant (k+1) a b$.

On the other hand, when $n=(k+1) a b$, the equation $a x+b y=n$ has exactly $k$ different positive integer solutions
$$
(x, y)=(j b, (k+1-j) a), j=1,2, \cdots, k,
$$

and when $n=(k-1) a b+a+b$, the equation $a x+b y=n$ also has exactly $k$ different positive integer solutions.
$$
(x, y)=(b j+1, a(k-1-j)+1), j=0,1, \cdots k
$$

Therefore, the maximum value of $n$ is $(k+1) a b$, and the minimum value is $(k-1) a b + a + b$.