7.2016 .

On the one hand, no two diagonals of the regular 2019-gon $\Gamma$ are perpendicular.

Assume there exist two diagonals of the regular 2019-gon $\Gamma$ that are perpendicular, say $AB \perp CD$, and $A, B, C, D$ are four vertices of $\Gamma$. Let $E$ be a point on the perpendicular bisector of $AB$, and let $E'$ be a point on the circumcircle of $\Gamma$ such that $EE'$ is its diameter.

Since $EC = E'D$, and $C, D, E$ are vertices of the regular 2019-gon $\Gamma$, it follows that $E'$ is also a vertex of $\Gamma$.

However, there cannot be two vertices of $\Gamma$ that are symmetric with respect to the center, so the assumption is false.

On the other hand, these non-intersecting diagonals can divide the regular 2019-gon $\Gamma$ into at most 2017 triangles, thus, at most 2016 diagonals can satisfy the condition. Furthermore, by selecting all the diagonals from a single vertex of $\Gamma$, the condition is met.
In summary, at most 2016 diagonals can satisfy the condition.
7. Given a regular 2019-gon, then, the maximum number of
$\qquad$ diagonals such that any two of them are either perpendicular or do not intersect except at endpoints is
