Examples of triangles, which related ellipses are perfectly packed with circles. Ellipse can be 
perfectly packed with $n$ circles
if 
\begin{align} 
b&=a\,\sin\frac{\pi}{2\,n}
\quad
\text{or equivalently, }\quad
e=\cos\frac{\pi}{2\,n}
,
\end{align} 
where $a,b$ are the major and minor semi-axis
of the ellipse and $e=\sqrt{1-\frac{b^2}{a^2}}$ is its eccentricity.
Consider a triangle and any ellipse, naturally associated with it,
for example, Steiner circumellipse/inellipse, Marden inellipse,
Brocard inellipse,
Lemoine inellipse,
ellipse with the circumcenter and incenter as the foci
and $r+R$ as the major axis, or any other ellipse
you can come up with, 
which can be consistently associated with the triangle.
The question is: provide the example(s) of triangle(s)
for which the associates ellipse(s) can be perfectly packed with circles.
Let's say that the max number of packed circles is 12,
unless you can find some especially interesting case with more circles.
For example, the Steiner incircle for the famous $3-4-5$ right triangle 
can not be perfectly packed. 
The example of the right triangle with the Marden inellipse,
perfectly packed with six circles is given in the self-answer below.


This is example of the 
ellipse, perfectly packed with three circles,
inscribed into the equilateral triangle $ABC$.
Let the center of the ellipse be $M=0$
and its semi-axes defined as
\begin{align}
s_a=|DF_1|=|DF_2|&=1
,\\
s_b=|MD|=|ME|&=\sin\frac\pi{2\cdot3}=\frac12
,
\end{align}
locations of the top and bottom points are
\begin{align}
D&=(0,-\tfrac12)
,\quad
E=(0,\tfrac12)
,\\
F_1&=(-\tfrac{\sqrt3}2,0)
,\quad
F_2=(\tfrac{\sqrt3}2,0)
.
\end{align}
\begin{align}
\text{Then the equation of the ellipse is }
\quad
x^2+2\,y^2&=0
\end{align}
and for the upper arc we have
\begin{align}
y(x)&=\tfrac12\,\sqrt{1-x^2}
,\\
y'(x)&=-\tfrac12\,\frac{x}{\sqrt{1-x^2}}
,
\end{align}
so ve can find the point $K$, tangent to 
the circumscribed equilateral $\triangle ABC$:
\begin{align}
-\tfrac12\,\frac{x}{\sqrt{1-x^2}}
&=\tan\tfrac\pi6=\tfrac{\sqrt3}3
,\\
x&=-\tfrac{2}{13}\sqrt{39}
,\\
y(x)&=\tfrac{1}{26}\sqrt{13}
.
\end{align}
So, the tangential points are
\begin{align}
K&=(-\tfrac{2}{13}\sqrt{39},\tfrac{1}{26}\sqrt{13} )
,\quad
L=(\tfrac{2}{13}\sqrt{39},\tfrac{1}{26}\sqrt{13} )
,
\end{align}
and the location of the vertices of $\triangle ABC$
can be easily found as
\begin{align}
A&=(-\tfrac16\,\sqrt3\,(\sqrt{13}+1), -\tfrac12)
,\quad
B=(\tfrac16\,\sqrt3\,(\sqrt{13}+1), -\tfrac12)
,\\
C&=(0,\tfrac12\,\sqrt{13})
,
\end{align}
the side length of the triangle is thus
\begin{align}
|AB|=|BC|=|CA|=a&=\tfrac13\,\sqrt3\,(\sqrt{13}+1)
,
\end{align}
and the tangential points $D,K,L$ divide the side segments 
as follows:
\begin{align}
|AK|=|BL|&=\sqrt{\tfrac2{39}(7+\sqrt{13})}
,\quad
|CK|=|CL|=\tfrac4{13}\,\sqrt{39}
.
\end{align} 
This ellipse is neither Steiner nor Mandart inellipse,
it is essentially a 
[![generalized Steiner inellipse]][Linfield1920],
for which the foci are 
the roots of the derivative of the rational
function
\begin{align}
f(z)&=(z-A)^u (z-B)^v (z-C)^w
,
\end{align}
where $A,B,C$ are the coordinates of the vertices of the triangle,
and the tangent points $L,K,D$ divide the segments
$BC,CA$ and $AB$
as $v:w,w:u$ and $u:v$, respectively.
In this case 
\begin{align}
u=v&=\tfrac1{12}\,(\sqrt{13}-1)
,\\
w&=\tfrac16\,(7-\sqrt{13})
.
\end{align}
Semi-axes of such ellipse, expressed in terms of the side length 
$a$ of the equilateral triangle, are
\begin{align}
s_a&=\frac{a\sqrt3}{12}\,(\sqrt{13}-1)
,\quad
s_b=\frac{a\sqrt3}{24}\,(\sqrt{13}-1)
.
\end{align} 
Similarly, for another orientation, 

we have
\begin{align}
s_a&=\frac{a\sqrt3}{3}\,(\sqrt{7}-2)
,\quad
s_b=\frac{a\sqrt3}{6}\,(\sqrt{7}-2)
\end{align} 
and 
\begin{align}
u&=\tfrac{11}3-\tfrac43\,\sqrt7
,\quad
v=\tfrac23\,\sqrt7-\tfrac43
,\quad
w=v
.
\end{align}
Reference
[Linfield1920]: Ben-Zion Linfield. 
“On the relation of the roots and poles of a rational function to the roots of its
derivative”. 
In: Bulletin of the American Mathematical Society 27.1 (1920), pp. 17-21 
