10.8. A straight rod 2 meters long was sawn into $N$ sticks, the length of each of which is expressed in whole 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?

(A. Magazinov)
Answer. $N=102$.

Solution. First solution. Let $N \leqslant 101$. Cut the stick into $N-1$ sticks of 1 cm each and one stick of $201-N$ cm. It is impossible to form a rectangle from this set, as each side of the rectangle is less than half the perimeter, and thus the stick of length $201-N \geqslant 100$ cm cannot be part of any side. Therefore, $N \geqslant 102$.

We will show that for $N=102$, the desired rectangle can be found. To do this, note that among all the sticks, there will be two sticks of 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, which is incorrect.

Set aside these two sticks. Let the lengths of the remaining sticks be $a_{1}, a_{2}, \ldots, a_{100}$ cm, then we have $a_{1} + a_{2} + \ldots + a_{100} = 198$. Among the 100 numbers $A_{1} = a_{1}, A_{2} = a_{1} + a_{2}, A_{3} = a_{1} + a_{2} + a_{3}, \ldots, A_{100} = a_{1} + a_{2} + \ldots + a_{100}$, there will be two that give the same remainder when divided by 99. Let these be $A_{k}$ and $A_{\ell}, k < \ell$. The number $A_{\ell} - A_{k}$ is strictly greater than zero and strictly less than 198, and it is divisible by 99. Thus, $A_{\ell} - A_{k} = 99 = a_{k+1} + a_{k+2} + \ldots + a_{\ell}$.

Thus, we have found several sticks with a total length of 99 cm. Set these aside as well. The remaining sticks also have a total length of 99 cm. Therefore, we can form a rectangle of $1 \times 99$ cm.

Second solution. We will provide another proof that for $N=102$, it is possible to form a rectangle.

Let the lengths of the sticks in the set, expressed in centimeters, be $a_{1}, a_{2}, \ldots, a_{102}$. We have $a_{1} + a_{2} + \ldots + a_{102} = 200$. Consider a circle of length 200 and divide it into 102 red points, forming arcs of lengths $a_{1}, a_{2}, \ldots, a_{102}$. These points are some 102 vertices of a regular 200-gon $T$ inscribed in this circle. The vertices of $T$ are paired into opposite pairs. There are 100 such pairs, and 102 red points, so among the red points, there will be two pairs of opposite points.

These two pairs of points divide the circle into two pairs of equal arcs. Thus, we have divided all the sticks into four groups $A, B, C, D$, where the total lengths in groups $A$ and $C$, as well as in groups $B$ and $D$, are equal. Therefore, it is possible to form a rectangle using each group to form one of its sides.

Comment. Only the correct answer - 0 points.

An example is provided showing that $N \geqslant 102 - 1$ point. It is proven that $N=102$ works, but its minimality is not justified - 5 points.