The pyramid must be a tetrahedron. Otherwise, the base wouldn't be a triangle, which would make the construction impossible. Let $D_0=s\sqrt{2}=2019$. Now consider the cube with edge length $s$, with four non-adjacent vertices being the vertices of the tetrahedron $T_1$. Consider the non-regular tetrahedrons with the base being an equilateral with side length $s\sqrt2$ and the other three faces being isosceles right triangles of leg length $s$, that are formed by choosing any vertex of the cube and the three vertices adjacent to that vertex. Now, take the midpoints of the base of the equilateral and connect them to form an equilateral triangle with side length $\frac{s\sqrt2}{2}$. Additionally, the segments connecting the midpoints of the base to the top of the tetrahedron also have length $\frac{s\sqrt2}{2}$, so a tetrahedron is formed. Repeat this for the other three vertices of the cube not adjacent to each other, and to the first vertex chosen. This will form four smaller tetrahedrons, one on each side of the original tetrahedron, which will form a second tetrahedron of the same size as the first. The vertices of the figure are the vertices of the cube, plus the intersection points of the tetrahedrons, which are the centers of the faces of the cubes. Similarly, to form $D_2$, we split the cube in half all three ways, forming three faces of a tetrahedron in each of the smaller cubes. We then inscribe a second tetrahedron with different vertices than the first in each of the smaller cubes. This forms vertices of the figure on the midpoints of the edges, and on the midpoints of the segments formed by the midpoints on adjacent edges on each face. In general, if we consider the vertices of $D_n$, they will be the alternating intersections of the segments that split each face of the cube into $2^n$ identical rows and $2^n$ identical columns (i.e. no two intersections are adjacent to each other, with the vertices of the cubes being intersections.) $D_1$ is the distance between two opposite vertices of the cube, which is just $s\sqrt{3}$. For $n>1$, we want to maximize the distance between two vertices, so we first take $V$, a vertex of the cube. Now we only need to consider the points closest to the opposite vertex. There are two distinct orientations for these points: $A$, $\frac{s}{2^{n-1}}$ units away from the opposite vertex, on the edge of the cube, and $B$, $\frac{\sqrt2s}{2^{n}}$ units away from the opposite vertex, on the diagonal of the face of the cube. But we notice that the plane through the diagonal $B$ is on and through the point $V$ is perpendicular to $AB$, and thus $VBA$ is a right angle and $VA>VB$. By Pythagorean Theorem in 3D the distance is thus $\sqrt{s^2+s^2+(s-\frac{s}{2^{n-1}})^2}=s\sqrt{3-\frac{1}{2^{n-2}}+\frac{1}{2^{2(n-1)}}}$ Since $s=\frac{2019\sqrt{2}}{2}$, $D_n=\left\{ \begin{array}{cl} 2019 & \text{if } n = 0, \\ \frac{2019\sqrt{6}}{2} & \text{if } n =1 , \\ \frac{2019\sqrt{6-\frac{1}{2^{n-3}}+\frac{1}{2^{2n-3}}}}{2} & \text{if } n > 1. \end{array} \right.$

$D_n=\left\{ \begin{array}{cl} 2019 & \text{if } n = 0, \\ \frac{2019\sqrt{6}}{2} & \text{if } n =1 , \\ \frac{2019\sqrt{2}}{2}\sqrt{3-2^{2-n}+ 2^{2-2n}} & \text{if } n > 1. \end{array} \right.$