Chekanov Yu.V.

A rectangle of size $1 \times k$ for any natural $k$ will be called a strip. For which natural $n$ can a rectangle of size $1995 \times n$ be cut into pairwise distinct strips?
Idea of the solution: we take the maximum strip (equal to the maximum side of the rectangle). The remaining strips will be combined in pairs, giving the sum of the maximum strip. If we have filled the rectangle, the problem is solved; otherwise, reasoning with areas shows that the rectangle cannot be cut into different strips.

Consider two cases: $n \leq 1995$ and $n > 1995$.

Case $n \leq 1995$. If $n \leq 998$, we cut the rectangle into $n$ strips of length 1995. We keep the first one, cut the second into two strips of length 1 and 1994, the third into two strips of length 2 and 1993, and so on. The last strip $1 \times 1995$ will be cut into parts $1 \times (n-1)$ and $1 \times (1996 - n)$. We get strips with lengths:

$$
\text { 1, 2,..., n - 1, } 1996 \text { - n, (1996 - n) + 1,..., 1994, } 1995.
$$

It is clear that the first $n-1$ strips are different, and the remaining strips are also different. However, to ensure that among the first $n-1$ strips there are no strips equal to any of the remaining strips, it is necessary (and sufficient) that $n - 1 < 1996 - n$, i.e., $n < 999$. The case $n = 1995$ is analogous. We cut the rectangle into 1995 strips of length $n$. We keep the first one, cut the second into two strips of length 1 and $n-1$, and so on. We get strips with lengths

$$
\text { 1, 2,..., 1994, } n-1994, (n-1994)+1, \ldots, n
$$

They will be different if $1994 < n - 1994$, i.e., $n \geq 3989$. To prove the necessity of this condition, we again compare the areas. Since the total area of the strips does not exceed $\frac{n(n+1)}{2}$, we get the inequality

$$
1995 n \leq \frac{n(n+1)}{2}
$$

From this, $n \geq 2 \cdot 1995 - 1 = 3989$.

Comments. $1^{0}$. A rectangle $n \times m$ with $n \geq m$ can be cut into strips of different lengths if and only if $n \geq 2m - 1$.

$2^{0}$. We used the formula $1+2+\ldots+n=\frac{n(n+1)}{2}$. This formula can be proven by induction. Moreover, it is a special case of the formula for the sum of an arithmetic progression. Nevertheless, we will provide an elegant proof of this formula. Let $1+2+\ldots+n=X$ and calculate the sum of all numbers in the table:

$$
\begin{array}{cccccc}
1 & 2 & 3 & \ldots & n-2 & n-1 & n \\
n & n-1 & n-2 & \ldots & 3 & 2 & 1
\end{array}
$$

Since the sum in each column is $n+1$ and there are $n$ columns, the sum of all numbers in the table is $n(n+1)$. On the other hand, the sum in each row is $X$, so the sum of all numbers in the table is $2X$. Therefore, $2X = n(n+1)$, from which our statement follows.