4. Given that the area of $\triangle A B C$ is $S$, the three medians of $\triangle A B C$ form $\triangle A_{1} B_{1} C_{1}$, with area $S_{1}$, and the three medians of $\triangle A_{k} B_{k} C_{k}(k=1,2, \cdots)$ form $\triangle A_{k+1} B_{k+1} C_{k+1}$, with area $S_{k+1}$. Then
$$
S_{1}+S_{2}+\cdots+S_{n}+\cdots=
$$
$\qquad$
4.3S.

As shown in Figure 2, the three medians of $\triangle ABC$ intersect at point $G$. Extend $AD$ to point $M$ such that $DM = DG$. Connect $BM$ and $CM$. Then quadrilateral $BMC G$ is a parallelogram. Therefore,
$$
\frac{BM}{CF} = \frac{GM}{AD} = \frac{BG}{BE} = \frac{2}{3}.
$$

Hence, $\triangle BGM$ is similar to $\triangle A_1B_1C_1$, and the similarity ratio is $\frac{2}{3}$. Therefore, $S_1 = S_{\triangle A_1B_1C_1} = \frac{9}{4} S_{\triangle BGM} = \frac{3}{4} S$.
Consequently, the sequence $\{S_n\}$ is a geometric sequence with the first term $S_1 = \frac{3}{4} S$ and common ratio $\frac{3}{4}$. Thus,
$$
S_1 + S_2 + \cdots + S_n + \cdots = \frac{\frac{3}{4} S}{1 - \frac{3}{4}} = 3S.
$$