Answer: 132

Solution:

Since $\lfloor x \rfloor$ and $\lfloor y \rfloor$ are integers whose squares sum to $25,$ the pair $(\lfloor x \rfloor, \lfloor y \rfloor)$ is one of the $12$ pairs $(\pm 5, 0),$ $(0, \pm 5),$ $(\pm 3, \pm 4),$ $(\pm 4, \pm 3).$ So $\mathcal{R}$ is the union of the $12$ unit squares whose lower-left corners are these points.

Let $K$ be the closure of $\mathcal{R}.$ Any closed disk containing $\mathcal{R}$ also contains $K,$ so the two sets have the same minimum enclosing radius. The map $(x, y) \mapsto (1 - x, 1 - y)$ permutes the closed unit squares in $K,$ so $K$ is symmetric under $180^\circ$ rotation about $Q = \left(\frac{1}{2}, \frac{1}{2}\right).$ If $X \in K,$ its opposite point $X' = 2Q-X$ also lies in $K.$ Every disk containing both endpoints of $\overline{XX'}$ has radius at least $\frac{XX'}{2} = XQ.$ Thus no enclosing disk can have radius smaller than the greatest distance from $Q$ to $K.$

That greatest distance is attained at square corners such as $A = (4, 5)$ and $B = (5, 4),$ and checking the corners of all twelve closed squares gives $$\begin{aligned} QA = QB &= \sqrt{\left(\tfrac{9}{2}\right)^2 + \left(\tfrac{7}{2}\right)^2} \\ &= \frac{\sqrt{130}}{2}. \end{aligned}$$ The disk centered at $Q$ with this radius contains every square, so it attains the lower bound.

Hence the minimum radius is $r = \frac{\sqrt{130}}{2},$ and $m + n = 130 + 2 = 132.$