$2n$ cells were marked on an infinite triangular grid, can you always find a triangle that contains exactly $n$ marked cells? The problem:
(Solve the problem for each positive integer $n$ separately)
$2n$ cells were marked on an infinite triangular grid, is it always possible to find a triangle (made by the grid lines) that contains exactly $n$ marked cells?
My progress:
I managed to solve the problem for every odd integer and for $n=4,6,8,10,12$
image links: all odd examples and
$4,6,8,10,12$
My inspiration to create the problem:
my inspiration
 A: Proof for $n$ odd
Consider $4k+2$ marked cells arranged as per the OP's diagrams and suppose that triangle $ABC$ contains exactly $2k+1$ marked cells, where $AB$ is horizontal.
The lines $AB$ and $AC$ divide the plane into $4$ regions. Let the region containing triangle $ABC$ contain exactly $2k+1+a$ marked cells and let the directly opposite region contain $\alpha$ marked cells. Define $b,c,\beta, \gamma$ similarly and then $$a+b+c+\alpha+\beta+\gamma=2k+1. ( 1 )$$
Considering edge $BC$ we see that $a+\beta+\gamma$ is even. Similarly, $b+\alpha+\gamma$ is even and $c+\alpha+\beta$ is $0,2k-2$ or $2k$.
If $\gamma \ne 0$, then all the marked points in triangle $ABC$ are in a triangle of side $3$ units which is clearly impossible. Therefore $\gamma = 0$.
If $\alpha +\beta=0$ then $a,b,c$ are all even, contradicting equation $1$. So there are two cases:-
If $c+\alpha+\beta=2k$. Then $a+b=1$. Considering edge $BC$ we see that $a$ is $0$ or at least $2$. The same is true for $b$ and so this case is impossible.
If $c+\alpha+\beta=2k-2$. Then $a+b=3$. Considering edge $BC$ again we see that $a$ is $0,2$ or at least $4$. The same is true for $b$ and so this case is also impossible.
