8. Given that a series of non-overlapping disks with radius $R$ can cover all lattice points (in the Cartesian coordinate system, points with both integer horizontal and vertical coordinates are lattice points) on the plane. Then $R_{\max }$ $\qquad$ 4 (fill in “greater than”, “less than”, or “equal to”).


8. Less than.

Place three disks $G\left(O_{i}, R\right)(i=1,2,3)$, each with radius $R$, such that they are pairwise externally tangent within a grid where the vertices are lattice points. Then, there must be a unit grid whose vertices $A, B, C, D$ are each contained in one of the three disks.

[Note] If a unit grid's vertex is in four disks, then $R \leqslant 1+\frac{\sqrt{2}}{2}$.

Without loss of generality, let $A, B \in G\left(O_{1}, R\right), C \in G\left(O_{2}, R\right), D \in G\left(O_{3}, R\right)$ (as shown in Figure 1).

Let the three points of tangency of $\odot O_{i}(i=1,2,3)$ be $E, F, G$. Without loss of generality, assume point $D$ is closer to $EF$ than $C$.

Draw a line through $D$ parallel to $EF$ intersecting the arcs $\overparen{FG}$ and $\overparen{EG}$ of $\odot O_{2}$ and $\odot O_{3}$ at points $C'$ and $D'$, respectively. Draw perpendiculars from $C'$ and $D'$ to $EF$ intersecting the arc $\overparen{EF}$ of $\odot O_{1}$ at points $B'$ and $A'$, respectively.
Then, quadrilateral $A'B'C'D'$ is a rectangle, and
$$
C'D' = 2a \leqslant 1, \quad A'D' = b \leqslant 1.
$$

Draw lines $l_{1} \parallel l_{2} \parallel O_{1}G$, and let the distance from $l_{i}(i=1,2)$ to $O_{1}G$ be $\frac{1}{2}$.

Let $l_{1}$ intersect the arcs $\overparen{EF}$ and $\overparen{EG}$ at points $A''$ and $D''$, and $l_{2}$ intersect the arcs $\overparen{EF}$ and $\overparen{FG}$ at points $B''$ and $C''$. Then,
$$
\begin{array}{l}
A''D'' = B''C'' \leqslant b \leqslant 1. \\
\text{Hence, } \sqrt{R^{2}-\left(R-\frac{1}{2}\right)^{2}}+\sqrt{R^{2}-\left(\frac{1}{2}\right)^{2}}+1 \geqslant \sqrt{3} R,
\end{array}
$$

i.e., $\sqrt{R-\frac{1}{4}}+\sqrt{R^{2}-\frac{1}{4}} \geqslant \sqrt{3} R-1$.
However, when $R=4$, inequality (1) does not hold, i.e., when $R=4$, it is impossible to cover all lattice points, and when $R>4$, the gap (the region $\triangle EFG$) in Figure 1 expands proportionally, thus it is also impossible to cover all lattice points.