3. A rectangle of size $1 \times k$ for any natural number $k$ will be called a strip. For which natural numbers $n$ can a rectangle of size $1995 \times n$ be cut into pairwise distinct strips?
3. Idea of the solution: 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: $n \leqslant 1995$ and $n>1995$.

For the case $n \leqslant 1995$. If $n \leqslant 998$, we will cut the rectangle into $n$ strips of length 1995. We keep the first one, cut the second one into two strips of length 1 and 1994, the third one 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 will get strips with lengths:

$$
1,2, \ldots, n-1,1996-n,(1996-n)+1, \ldots, 1994,1995
$$

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 one of the remaining strips, it is necessary (and sufficient) that $n-1 \leqslant 997$. This condition is satisfied for $n \leqslant 998$.

For the case $n \geqslant 1995$. The case $n \geqslant 1995$ is analogous. We cut the rectangle into 1995 strips of length $n$. We keep the first one, cut the second one into two strips of length 1 and $n-1$, and so on. We will get strips with lengths:

$$
1,2, \ldots, 1994, n-1994,(n-1994)+1, \ldots, n
$$

They will be different if $1994 < n-1994$, i.e., $n \geqslant 3989$. 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

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

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

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

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

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 $2 X$. Therefore, $2 X=n(n+1)$, from which our statement follows.