Mathematical Olympiad Rioplatense

Let $l$ and $m$ be grid lines in the initial rectangle that were cut by the first cutting. There was a folding that made them coincide, hence $l\parallel m$ (a folding cannot make perpendicular lines coincide). Also it follows that $l$ and $m$ are separated by at least one line of folding which is intact. So the grid lines cut by one cutting are all parallel, interior and no two of them are adjacent. Let their direction be perpendicular to a side of odd length $d$. Of the $d-1$ interior grid lines in this direction at most $\frac{d-1}{2}$ can be cut. So the cutting yields at most $\frac{d-1}{2}+1=\frac{d+1}{2}$ pieces. For $d=59$ we obtain $29$ lines and $30$ pieces. The same reasoning applies to the second cutting. It must be made in the other direction to reach a maximum number of parts. At most $\frac{133-1}{2}=66$ grid lines can be cut, so each piece from the first cutting decomposes into at most $67$ parts. Thus the total number of parts does not exceed $30\cdot 67=2010$.

For an example with $2010$ parts label $1,2,\dots ,132$ the interior grid lines parallel to side $59$. Fold along every line with odd label in a bandoneón-like fashion to obtain a rectangle $2\times 59$. Do the same in the other direction. Flattening down gives a $2\times 2$ square whose middle lines contain all grid lines that are not lines of folding. Two cuts along the middle lines yield $(66+1)(29+1)=2010$ parts.