Solution:

If $n$ is odd, in every row there is a rod of length 1 and therefore the number of rods of length 1 is at least $2(n+1)$; hence at least $n+1$ cuts are necessary. For the construction shown alongside in the case $n=7$, but easily generalizable to every odd $n$, exactly $n+1$ cuts are needed, which is therefore the minimum for odd $n$.

Let us now analyze the case of even $n$. Let $R_{2}$ be the second row and $C_{2}$ the second column of the grid. For every rod of length two that appears

in $R_{2}$, the column corresponding to the center of that rod must begin with a rod of length 1, and therefore, since the length of every column is even, it must have at least two rods of length 1.

So if we denote by $i$ the number of rods of length 2 that appear in $R_{2}$, there must be $n-2i$ rods of length 1 in $R_{2}$ and at least $2i$ rods of length 1 in the columns. In conclusion, between the second row and all the columns there are at least $n$ rods of length 1. Repeating the same reasoning for $C_{2}$, we find that there are at least $n$ rods of length 1 between $C_{2}$ and all the rows. Adding these two estimates, we find that the number of rods of length 1 is at least $2n-k$ where $k$ is the number of rods that have been counted twice (that is, those belonging to $C_{2}$ or to $R_{2}$).

In the estimate, two rods on $C_{2}$ (resp. $R_{2}$) are counted only if $R_{2}$ (resp. $C_{2}$) begins with a rod of length 2 (the rods would cross!). But this cannot happen simultaneously for $R_{2}$ and $C_{2}$, and therefore $k$ is at most 2. The minimum number of rods of length 1 is therefore $2n-2$, which forces $n-1$ cuts. The construction alongside achieves this value for $n=6$.