Three, let $a$ and $b$ be the lengths of the two legs of a right triangle, then the length of the hypotenuse $c = \sqrt{a^2 + b^2}$. Since $a$, $b$, and $c$ are all positive integers, $a \neq b$. Without loss of generality, assume $a > b$. According to the problem, we have
$$a + b + \sqrt{a^2 + b^2} = \frac{ab}{2}.$$

Squaring both sides and simplifying, we get $\frac{a^2 b^2}{4} - a^2 b - ab^2 + 2ab = 0$, which simplifies to $ab - 4a - 4b + 8 = 0$.
Thus, $(a-4)(b-4) = 8 = 1 \times 8 = 2 \times 4$.
Since $a$ and $b$ are positive integers and $a > b$, we have
$$\left\{\begin{array}{l} a - 4 = 8, \\ b - 4 = 1 \end{array} \text { or } \left\{\begin{array}{l} a - 4 = 4, \\ b - 4 = 2 \end{array}\right.\right.$$

Solving these, we get $a = 12, b = 5, c = 13$; or $a = 8, b = 6, c = 10$.
Therefore, the lengths of the sides of the right triangle are $(12, 5, 13)$ or $(8, 6, 10)$.