Well, you can use the same formulas for these pyramids as for the normal polyhedrons. So, initially we can see that you can't just multiply the diameters of spheres by the number of layers, but we can use some simple trigonometry.
The pyramid $\displaystyle S_{n} $ outlines a square pyramid, where its slant height is the diameter of a sphere multiplied by the number of layers. To find the height of the pyramid, you can use the Pythagorean Theorem to find how height is related to base length and slant height.
Here's a way to derive the formula for height from slant height:
Let $\displaystyle b $ equal the side length of the base of the pyramid.
Let $\displaystyle h $ equal the height of the pyramid.
Let $\displaystyle s $ equal the slant height of the pyramid.
$\displaystyle s^2=(\frac{b}{2})^2 + h^2 $
$\displaystyle s^2=\frac{b^2}{4} + h^2 $
$\displaystyle s^2-\frac{b^2}{4}= h^2 $
$\displaystyle h = \sqrt{s^2-\frac{b^2}{4}} $
And now for $\displaystyle T_{n} $; this sphere pyramid outlines a regular tetrahedron. All of the edge lengths are equal.
Considering that $\displaystyle a $ is the edge length of the tetrahedron, which is determined by multiplying the diameter of your sphere by the number of the spheres that line up along the edge of the pyramid, we can figure that $\displaystyle h $, the height of the pyramid is equal to $\displaystyle \frac{a\sqrt{6}}{3} $ by using some simple trig of 30-60-90 triangles.
So what you're essentially left with is $\displaystyle h_{S} - h_{T} > 2019 $, or by substitution:
$\displaystyle \sqrt{s^2-\frac{b^2}{4}}-\frac{a\sqrt{6}}{3} > 2019 $.
I am not going to immediately solve the entire problem for you, but I really hope this helps, and if you have any questions about what to substitute in for $\displaystyle a $, $\displaystyle b $, or $\displaystyle s $, or anything in general, please feel free to ask!
-Ivan