4. Let the edge length of the regular tetrahedron $ABCD$ be 2. Then the sum of the areas of all sections of the circumsphere of this tetrahedron by planes that are equidistant from the four vertices of the tetrahedron is $\qquad$
4. $\frac{59 \pi}{6}$.

There are two types of planes that meet the requirements.
(1) Planes parallel to one of the triangular bases, a total of 4.

It is easy to know that the distance between this type of section and the parallel base is $\frac{\sqrt{6}}{3}$, the distance from the center of the sphere to the base is $\frac{\sqrt{6}}{6}$, so the distance from the center of the sphere to the section is $\frac{\sqrt{6}}{6}$, and the radius of the circumscribed sphere is $\frac{\sqrt{6}}{2}$.
Therefore, the area of this section circle is $\pi\left(\frac{3}{2}-\frac{1}{6}\right)=\frac{4 \pi}{3}$.
(2) Planes parallel to a pair of opposite edges, a total of 3.
As shown in Figure 2, plane $E F G H$ is one of them, with the midpoint of $B D$ being $M$, and $N, L$ being the intersection points of plane $A M C$ with $E H, F G$ respectively.
It is easy to know that $B D \perp$ plane $A M C$.
Since $E H / / B D$, thus,
plane $E F G H \perp$ plane $A M C$,
$N L$ is their intersection line.
As shown in Figure 3, construct plane $A M C, A P \perp C M$ at point $P$, and $A P$ intersects $N L$ at point $O$.
Then $M A=M C=\sqrt{3}, A C=2$.
By $M P=\frac{\sqrt{3}}{3}, M L=\frac{\sqrt{3}}{2}$
$\Rightarrow L P=\frac{\sqrt{3}}{6}$.
It is easy to know that $\tan C=\sqrt{2}$.
Thus, $O P=L P \tan C=\frac{\sqrt{6}}{6}$.
Therefore, $O$ is the center of the circumscribed sphere, which is precisely on plane $E F G H$.
Hence, the area of this section circle is $\frac{3 \pi}{2}$.
In summary, the total area of the sections is
$$
\frac{4 \pi}{3} \times 4+\frac{3 \pi}{2} \times 3=\frac{59 \pi}{6} .
$$