Solution:
1. It is not difficult to see that the rectangles must be placed one over another so that any of them has two vertices on the sides $AC$ and $BC$, and the first one has a base on the side $AB$.
Now we shall prove by induction that the sum of the areas of $n$ such rectangles is maximal when the side $AC$ is divided into $n+1$ equal parts by the vertices of the rectangles lying on it. Then the sum equals $\frac{n}{n+1} S$, where $S = S_{ABC}$.
Let $MNPQ$ be a rectangle with $M, N \in AB$, $P \in BC$, $Q \in AC$. Setting $\frac{CQ}{AC} = x$, it easily follows that
$$S_{MNPQ} = 2x(1-x) S$$
Hence $S_{MNPQ}$ is maximal if $x = \frac{1}{2}$, i.e., when $Q$ is the midpoint of the side $AC$. This proves our statement for $n=1$.
Assume that the statement holds true for some $k$ and consider $k+1$ rectangles $M_i N_i P_i Q_i$ with $M_i, N_i \in P_{i-1} Q_{i-1}$ ($P_0 \equiv A, Q_0 \equiv B$) and $P_i \in BC$ and $Q_i \in AC$, $i=1,2, \ldots, k+1$. Setting $\frac{CQ_1}{AC} = x$, we get $S_{M_1 N_1 P_1 Q_1} = 2x(1-x) S$. The induction assumption implies that $\sum_{i=2}^{k+1} S_{M_i N_i P_i Q_i}$ is maximal if $Q_1 Q_2 = Q_2 Q_3 = \cdots = Q_k Q_{k+1} = Q_{k+1} C$. Therefore
$$\sum_{i=2}^{k+1} S_{M_i N_i P_i Q_i} \leq \frac{k S_{Q_1 P_1 C}}{k+1} = \frac{k x^2 S}{k+1}$$
Then
$$\begin{aligned} \sum_{i=1}^{k+1} S_{M_i N_i P_i Q_i} & \leq \left(2x(1-x) + \frac{k x^2}{k+1}\right) S \\ & = \left[\frac{k+1}{k+2} - \frac{k+2}{k+1}\left(x - \frac{k+1}{k+2}\right)^2\right] S \\ & \leq \frac{k+1}{k+2} S \end{aligned}$$
The equality is attained if $x = \frac{k+1}{k+2}$, i.e., when the points $Q_1, Q_2, \ldots, Q_{k+1}$ divide the side $AC$ into equal parts.