Answer: 56

Solution:

The argument of $\frac{z_3 - z_1}{z_2 - z_1}$ is the angle $\angle z_2 z_1 z_3,$ and the argument of $\frac{z - z_2}{z - z_3}$ is the angle between $\overline{zz_2}$ and $\overline{zz_3}.$ Their product is real exactly when these angles are equal or supplementary, which by the inscribed angle theorem happens exactly when $z_1,$ $z_2,$ $z_3,$ and $z$ are concyclic. So $z$ lies on the circumcircle of $z_1, z_2, z_3.$

The segment from $18 + 39i$ to $18 + 83i$ is vertical, so its perpendicular bisector is the horizontal line $y = 61.$ The segment from $z_2 = 18 + 39i$ to $z_3 = 78 + 99i$ has slope $1$ and midpoint $(48, 69),$ so its perpendicular bisector is $y - 69 = -(x - 48).$ Setting $y = 61$ gives $x = 56,$ so the center is $56 + 61i.$

The point of the circle with maximal imaginary part is directly above the center, so the real part of $z$ is $56.$