Problem 1. Cut 2003 rectangles from an acute $\triangle A B C$ such that any of them has a side parallel to $A B$ and the sum of their areas is maximal.
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 $A C$ and $B C$, and the first one has a base on the side $A B$.

Now we shall prove by induction that the sum of the areas of $n$ such rectangles is maximal when the side $A C$ 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 $S=S_{A B C}$.

Let $M N P Q$ be a rectangle with $M, N \in A B, P \in B C, Q \in A C$. Setting $\frac{C Q}{A C}=x$, it easily follows that

$$
S_{M N P Q}=2 x(1-x) S
$$

Hence $S_{M N P Q}$ is maximal if $x=\frac{1}{2}$, i.e., when $Q$ is the midpoint of the side $A C$. 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 $M_{i}, N_{i} \in P_{i-1} Q_{i-1}\left(P_{0} \equiv A, Q_{0} \equiv B\right)$ and $P_{i} \in B C$ и $Q_{i} \in A C, i=1,2, \ldots, k+1$. Setting $\frac{C Q_{1}}{A C}=x$, we get $S_{M_{1} N_{1} P_{1} Q_{1}}=2 x(1-x) S$. 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

$$
\begin{aligned}
\sum_{i=1}^{k+1} S_{M_{i} N_{i} P_{i} Q_{i}} & \leq\left(2 x(1-x)+\frac{k x^{2}}{k+1}\right) S \\
& =\left[\frac{k+1}{k+2}-\frac{k+2}{k+1}\left(x-\frac{k+1}{k+2}\right)^{2}\right] S \\
& \leq \frac{k+1}{k+2} S
\end{aligned}
$$

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 $A C$ into equal parts.