Assume you are building towers out of identical frictionless cubes of constant density on a flat frictionless surface. The rules of the tower are that there may only be a single cube that does not support any other cubes (only one cube on the top layer, all other cubes support that top cube), if any cube other than this top cube is removed the tower will topple (the tower is minimal), and the supporting cubes may not be the only support for another cube (lower levels must be larger than higher levels). These rules imply that if one is required to use $n$ cubes, not all values of $n$ permit an acceptable tower. The case of $n=1,3,4,6$ are all possible, as demonstrated in the image below. The cases of $n=2,5$ are not possible (I am fairly certain). My question is for what values of $n$ may one build an acceptable tower?
Only $2$ and $5$ are not possible.
Clearly every triangular number is possible. You can build a triangle just like your $n=6$ example, where each layer is a row that contains one fewer cubes than the row below it.
In such a triangular tower with 3 or more blocks on the bottom row, any of the bottom row blocks can be replaced by two blocks. If the bottom row block being replaced is in the middle supporting two blocks above it, then place the replacement blocks such that each supports only one of the two blocks at a corner. That ensures minimality. Once all of the middle blocks of the bottom row are replaced by two, you can replace each end block by two arranged just like the $n=4$ case example. In this way you can add blocks one by one until the entire bottom row is doubled. If you then add one more cube you have reached the next triangular number, so you can put all those extra cubes and your new cube together to form a new base row.
Here is how it looks when a base row of 5 blocks supporting a row of 4 blocks is doubled to 10 blocks:
This procedure works for any triangular tower with 3 or more blocks on the bottom row, so from $6$ cubes onwards.
As Oscar Lanzi points out, this does not work on a tower with only 2 blocks in the bottom row, because then you would have a single block being supported by 4 others, which cannot be minimal. So the only numbers that are not possible are $2$ and $5$.