4. If the three sides of a right-angled triangle are positive integers, and the numerical value of its perimeter is equal to the numerical value of its area, such a right-angled triangle is called "harmonious." Then, the sum of the radii of the circumcircles of all "harmonious" right-angled triangles is $\qquad$

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4. $\frac{23}{2}$.

Analysis: Let the legs of the right triangle be $a, b$. According to the problem, we have $a+b+\sqrt{a^{2}+b^{2}}=\frac{a b}{2}$. Rearranging and squaring, we get $\frac{a^{2} b^{2}}{4}+2 a b-a b(a+b)=0$, which simplifies to $a+b-\frac{a b}{4}=2$. Therefore, $(a-4)(b-4)=8$, yielding $a=5,6,8,12$, and the corresponding $b=12,8,6,5$. Thus, there are two triangles that satisfy the conditions, with side lengths $(6,8,10)$ and $(5,12,13)$. Therefore, the sum of the circumradii of the right triangles that satisfy the condition is $\frac{23}{2}$.