Three. (20 points) A bag contains $m$ red balls and $n$ white balls $(m > n \geqslant 4)$, which are identical except for their colors. Now, two balls are drawn at random.
(1) If the probability of drawing two red balls is an integer multiple of the probability of drawing one red and one white ball, prove that $m$ must be odd;
(2) If the probability of drawing two balls of the same color is equal to the probability of drawing two balls of different colors, find all pairs $(m, n)$ that satisfy $m+n \leqslant 40$.
(1) Let "drawing two red balls" be event $A$, and "drawing one red and one white ball" be event $B$. Then,
$$
P(A)=\frac{\mathrm{C}_{m}^{2}}{\mathrm{C}_{m+n}^{2}}, P(B)=\frac{\mathrm{C}_{m}^{1} \mathrm{C}_{n}^{1}}{\mathrm{C}_{m+n}^{2}}.
$$

According to the problem, $P(A)=k P(B)\left(k \in \mathbf{N}_{+}\right)$. Therefore,
$$
\mathrm{C}_{m}^{2}=k \mathrm{C}_{m}^{1} \mathrm{C}_{n}^{1}.
$$

From this, we get $m=2 k n+1$.
Since $k, n \in \mathbf{N}_{+}$, $m$ is an odd number.
(2) Let "drawing two white balls" be event $C$, then
$$
P(C)=\frac{\mathrm{C}_{n}^{2}}{\mathrm{C}_{m+n}^{2}}.
$$

According to the problem, $P(A)+P(C)=P(B)$. Therefore,
$$
\mathrm{C}_{m}^{2}+\mathrm{C}_{n}^{2}=\mathrm{C}_{m}^{1} \mathrm{C}_{n}^{1}.
$$

From this, we get $m+n=(m-n)^{2}$.
Thus, $m+n$ is a perfect square.
Given $m>n \geqslant 4$ and $m+n \leqslant 40$, we have $9 \leqslant m+n \leqslant 40$.
Therefore,
$$
\left\{\begin{array}{l}
m+n=9, \\
m-n=3 ;
\end{array}\left\{\begin{array}{l}
m+n=16 \\
m-n=4 ;
\end{array}\right.\right.
$$
$$
\left\{\begin{array}{l}
m+n=25, \\
m-n=5 ;
\end{array} \quad \left\{\begin{array}{l}
m+n=36 \\
m-n=6 .
\end{array}\right.\right.
$$

Hence, the tuples $(m, n)$ that satisfy the conditions are
$$
(10,6),(15,10),(21,15).
$$