Magozinov A.

A straight stick 2 meters long was sawn into $N$ sticks, the length of each of which is expressed as an integer number of centimeters. For what smallest $N$ can it be guaranteed that, using all the resulting sticks, one can, without breaking them, form the contour of some rectangle?
Let $N \leq 101$. We cut the stick into $N-1$ sticks of length 1 cm and one stick of length $201-N$ cm. It is impossible to form a rectangle from this set because each side of the rectangle must be less than half the perimeter, and thus the stick of length $201-N \geq 100$ cm cannot be part of any side. Therefore, $N \geq 102$.

We will show that for $N=102$, the desired rectangle can be formed.

First method. Notice that among all the sticks, there will be two sticks of length 1 cm each. Indeed, if this were not the case, the total length of the sticks would be at least $2 \cdot 101 + 1 = 203$ cm.

Set aside these two sticks. Let the lengths of the remaining sticks be $a_{1}, a_{2}, \ldots, a_{100}$ cm, with $a_{1} + a_{2} + \ldots + a_{100} = 198$. Among the 100 numbers $A_{1} = a_{1}$, $A_{2} = a_{1} + a_{2}$, $A_{3} = a_{1} + a_{2} + a_{3}$, ..., $A_{100} = a_{1} + a_{2} + \ldots + a_{100}$, there will be two that give the same remainder when divided by 99. Let these be $A_{k}$ and $A_{l}$, with $k < l$. Then $A_{l} - A_{k} = a_{k+1} + a_{k+2} + \ldots + a_{l} = 99m$ for some integer $m$. 

If $m=1$, we have found a segment of length 99. If $m>1$, we can remove the last laid stick of length $m$ and replace it with $99 - L$ unit sticks (this is possible because $99 - L < m \leq l$). Now we have laid three sides of the rectangle (1, 1, and 99). Laying the remaining sticks in a row, we get another side of length 99.

## Answer

For $N=102$.