Three, (25 points) In $\triangle A B C$, $C D \perp A B$ intersects $A B$ at point $D, R$ and $S$ are the points where the incircles of $\triangle A C D$ and $\triangle B C D$ touch $C D$. If $A B$, $B C$, and $C A$ are three consecutive positive integers, and $R S$ is an integer, find the value of $R S$.

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In $\triangle A B C$, $C D \perp A B$ intersects $A B$ at point $D, R$ and $S$ are the points where the incircles of $\triangle A C D$ and $\triangle B C D$ touch $C D$. If $A B$, $B C$, and $C A$ are three consecutive positive integers, and $R S$ is an integer, find the value of $R S$.
Three, as shown in Figure 9, let the sides of $\triangle ABC$ be $BC = a$, $CA = b$, and $AB = c$. Let $CD = h$, $AD = x$, and $BD = y$. The radii of the two incircles are $r_1$ and $r_2$. Then,
$$
\begin{array}{l}
RS = |RD - SD| \\
= |r_1 - r_2|. \\
\text{Since } b = AC = AP + CP = AE + CR \\
= (x - r_1) + (h - r_1),
\end{array}
$$

Thus, $r_1 = \frac{x + h - b}{2}$.
Similarly, $r_2 = \frac{y + h - a}{2}$.
Therefore, $|RS| = |r_1 - r_2|$
$$
\begin{array}{l}
= \left| \frac{x + h - b}{2} - \frac{y + h - a}{2} \right| \\
= \frac{1}{2} |(x - y) + (a - b)|.
\end{array}
$$

Also, $a^2 - y^2 = h^2 = b^2 - x^2$, so $x^2 - y^2 = b^2 - a^2$.
Since $x + y = c$, we have
$$
x - y = \frac{(b + a)(b - a)}{c}.
$$

Substituting equation (2) into equation (1) gives
$$
\begin{array}{l}
RS = \frac{1}{2} \left| \frac{(b + a)(b - a)}{c} + (a - b) \right| \\
= \frac{|b - a|}{2c} |a + b - c|.
\end{array}
$$

Since $a$, $b$, and $c$ are three consecutive natural numbers, let these three consecutive natural numbers be $n$, $n+1$, and $n+2$.

Since equation (3) is symmetric with respect to $a$ and $b$, we can assume without loss of generality that $a > b$. There are three cases:
(1) $a = n+1$, $b = n$, $c = n+2$, in this case, $RS = \frac{1 \times (n-1)}{2(n+2)}$ is not an integer.
(2) $a = n+2$, $b = n$, $c = n+1$, in this case, $RS = \frac{2(n+1)}{2(n+1)} = 1$, there are infinitely many such $n$, so there are infinitely many triangles that satisfy the condition.
(3) $a = n+2$, $b = n+1$, $c = n$, in this case, $RS = \frac{n+3}{2n}$, for $RS$ to be a positive integer, then $n+3 \geq 2n$.
Thus, $n \leq 3$.
Since $n$ is a positive integer, then $n = 1, 2, 3$.
When $n = 1$, $a = 3$, $b = 2$, $c = 1$, in this case, $b + c = a$ cannot form a triangle, so $n \neq 1$.
When $n = 2$, $RS = \frac{5}{4}$ is not an integer.
When $n = 3$, $RS = 1$.
In summary, $RS = 1$.