Let represent the rod with the $ [0,200] $ interval. Then every cut is some point between $1,2,...,199$. If $ N \le 101 $ it is impossible to guarantee rectangle's existence, e.g. let cuts be on points: 1,2,...,100. When $N=102$, such rectangle exists. Proof: Lets consider pairs: $(i, 100+i), \, i=1,2,\dots,99 $ and point $100$. Their union cover all integer points inside [0,200] . Because we have $ 101$ cuts, then there are two cuts on the same pair $(i, 100+i)$. Let put together sticks $[0,i]$ and $[100+i, 200]$ and let denote the formed rod as $A$. Let $B$ denote the rod $[i, 100+i]$. Both intervals $A$ and $B$ have length $100$. We have one cut $i$ on $A$, there remain other $99$ cuts(because we don't take in account cut $100+i $ on the original stick) It is clear that there exists $j \in \{1,2,\dots,99\}$, so that we have cut $j$ on both sticks $A$ and $B$. Then 4 rods: $[0, j],\, [j,100]$ on $A$ and $[0, j],\, [j,100]$ on $B$ will form a rectangle as required.