Answer: $N = 102$.

First solution. Suppose $N \leq 101$. Cut the stick into $N-1$ pieces of length $1$ cm and one piece of length $(201 - N)$ cm. From this set, it is impossible to form a rectangle, since each side of the rectangle is less than the semiperimeter, and therefore the piece of length $201 - N \geq 100$ cm cannot be part of any side. Thus, $N \geq 102$.

Let us show that for $N = 102$ it is possible to form a rectangle. For this, note that among all the pieces there will be two of length $1$ cm. Indeed, if this were not the case, the total length of the pieces would be at least $2 \cdot 101 + 1 = 203$ cm, which is not true.

Set aside these two pieces. Let the lengths of the remaining pieces be $a_1, a_2, \dots, a_{100}$ cm, then $a_1 + a_2 + \dots + a_{100} = 198$. Among the $100$ numbers $A_1 = a_1, A_2 = a_1 + a_2, A_3 = a_1 + a_2 + a_3, \dots, A_{100} = a_1 + a_2 + \dots + a_{100}$, there will be two with 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 $99$. Therefore, $A_\ell - A_k = 99 = a_{k+1} + a_{k+2} + \dots + a_\ell$.

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

Second solution. Here is another proof that for $N = 102$ it is possible to form a rectangle.

Let the lengths of the pieces, in centimeters, be $a_1, a_2, \dots, a_{102}$. We have $a_1 + a_2 + \dots + a_{102} = 200$. Consider a circle of length $200$ and mark $102$ red points dividing it into arcs of lengths $a_1, a_2, \dots, a_{102}$. These points are some $102$ vertices of a regular $200$-gon $T$ inscribed in this circle. The vertices of $T$ are divided into pairs of opposite points. There are $100$ such pairs, and $102$ red points, so among the red points there are 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 pieces into four groups $A, B, C, D$, with the total lengths in groups $A$ and $C$, and in groups $B$ and $D$, being equal. Therefore, we can form a rectangle, using each group to make one side.

Third solution. Here is yet another proof that for $N = 102$ it is possible to form a rectangle. We show how to lay out a rectangle $1 \times 99$.

Let $\ell \geq 2$ be the largest among the lengths of all the pieces. Let $x$ be the number of pieces of length $1$. Then, besides these $x$ pieces and the piece of length $\ell$, there are $101 - x$ pieces, each of length at least $2$. Hence $\ell + x + 2(101 - x) \leq 200$, and $x \geq \ell + 2$. So, there are at least $\ell + 2$ pieces of length $1$. Set aside two pieces of length $1$—use them to make two sides of the rectangle, leaving $\ell$ pieces of length $1$.

Start laying out the pieces in order of decreasing length to form one side of length $99$. Suppose at some step the row has length $L < 99$, and after adding the next piece of length $m$ it becomes $L + m > 99$. Then remove the last piece of length $m$, and instead place $99 - L$ pieces of length $1$ (this can be done since $99 - L < m \leq \ell$). Now we have three sides of the rectangle ($1$, $1$, and $99$). Laying out the remaining pieces in a row, we get another side of length $99$.