Answer: $n \equiv 0,2(\bmod 6)$ We first prove that $n \equiv 0,2(\bmod 6)$ is necessary for there to be such a partitioning. We break this down into proving that $n$ has to be even and that $n \equiv 0,2(\bmod 3)$. The only way a segment on a side of the square can be part of such a T-shape is as one of the two consecutive segments along the longer side of the T-shape, so they must come in pairs and therefore, the length of each side has to be even. On the other hand, the total number of segments, which is $2 n(n+1)$, has to be a multiple of three as each T-shape consists of three segments, hence either $n$ or $n+1$ is a multiple of 3 , implying that $n \equiv 0,2(\bmod 3)$. We can then show that these two conditions is sufficient by showing that $n=2$ and $n=6$ works and $n=k+6$ works whenever $n=k$ works. The construction for $n=2$ is simple; just put a T-shape with the longer side on each of the four sides. For $n=6$ and to go from $n=k$ to $n=k+6$, consider the following diagram: There are two main parts - the cycle of stacks of T's in all four orientation (see the red, blue, yellow, and green stacks), and the border (seen here by the cyan, brown, and black T-shapes). The case $n=6$ can be considered as a special case where the middle square is a single point.