Starting from a pyramid $T_0$ whose edges are all of length $2019$, we construct the Figure $T_1$ when considering the triangles formed by the midpoints of the edges of each face of $T_0$, building in each of these new pyramid triangles with faces identical to base. Then the bases of these new pyramids are removed. Figure $T_2$ is constructed by applying the same process from $T_1$ on each triangular face resulting from $T_1$, and so on for $T_3, T_4, ...$ 

Let $D_0= \max \{d(x,y)\}$, where $x$ and $y$ are vertices of $T_0$ and $d(x,y)$ is the distance between $x$ and $y$. Then we define $D_{n + 1} = \max \{d (x, y) |d (x, y) \notin \{D_0, D_1,...,D_n\}$, where $x, y$ are vertices of $T_{n+1}$. 

 Find the value of $D_n$ for all $n$.
1. **Initial Setup and Definitions:**
   - We start with a pyramid \( T_0 \) whose edges are all of length \( 2019 \). This pyramid is a regular tetrahedron.
   - Define \( D_0 = \max \{d(x,y)\} \) where \( x \) and \( y \) are vertices of \( T_0 \) and \( d(x,y) \) is the distance between \( x \) and \( y \). For a regular tetrahedron with edge length \( s \), the maximum distance between any two vertices is the length of the edge, so \( D_0 = 2019 \).

2. **Constructing \( T_1 \):**
   - To construct \( T_1 \), we consider the midpoints of the edges of each face of \( T_0 \). These midpoints form a smaller tetrahedron inside \( T_0 \).
   - The edge length of this smaller tetrahedron is half the edge length of \( T_0 \), i.e., \( \frac{2019}{2} \).

3. **Distance Calculation for \( T_1 \):**
   - The maximum distance between any two vertices of \( T_1 \) is the same as the edge length of \( T_1 \), which is \( \frac{2019}{2} \).
   - However, we need to find the next maximum distance that is not already in the set \(\{D_0\}\). This distance is the space diagonal of the cube formed by the midpoints of \( T_0 \).
   - The space diagonal of a cube with edge length \( \frac{2019}{2} \) is given by \( \sqrt{3} \times \frac{2019}{2} = \frac{2019\sqrt{3}}{2} \).

4. **Generalizing for \( T_n \):**
   - For \( T_n \), the edge length of the tetrahedron formed at each step is \( \frac{2019}{2^n} \).
   - The maximum distance \( D_n \) is the space diagonal of the cube formed by the midpoints of the tetrahedron at step \( n \).
   - The space diagonal of a cube with edge length \( \frac{2019}{2^n} \) is \( \sqrt{3} \times \frac{2019}{2^n} = \frac{2019\sqrt{3}}{2^n} \).

5. **Formulating \( D_n \):**
   - For \( n = 0 \), \( D_0 = 2019 \).
   - For \( n = 1 \), \( D_1 = \frac{2019\sqrt{3}}{2} \).
   - For \( n > 1 \), \( D_n = \frac{2019\sqrt{3}}{2^n} \).

The final answer is \( \boxed{ D_n = \frac{2019\sqrt{3}}{2^n} } \) for all \( n \geq 1 \).