Solved with amar_04 and srijonrick Huge thanks to biomathematics for correcting few errors and checking the solution Solution : Replace $N$ with $n$ and call a number $n$ ngood if there exists no such equilateral triangle satisfying all the given conditions and good otherwise. Claim : For $n>3$ and $n$ even, is ngood. Proof : We first do the following $\quad$ - $\quad$ We divide the plane in two parts via a line, say $\ell$ which runs alongside grid lines. For brevity, let's assume this line to be horizontal and rotate the plane accordingly. It's clear that any unit triangle will be lying on only one side of $\ell$ as either above or below, and now we consider those $2$ black unit triangles, one with one side lying on this line $\ell$, and other, the reflected versions (of the same triangle) along the same line $\ell$. Call such pair nice and a configuration having nice pairs only, as nice configuration. Now since we are considering $n$ to be even, we will consider nice configurations. We want to show that we can't find $\mathcal{S}$ satisfying given properties. First let $p$ and $q$ be the properties of lines which are $60^\circ$ and $120^\circ$ anti-clockwise rotated from $\ell$ and let $\mathbb{P}$ and $\mathbb{Q}$ be their respective sets of such lines. Select a line $t$ from either $\mathbb{P}$ or $\mathbb{Q}$. Define right shifting or left shifting for $t$ as moving it $1$ step forward or backward respectively. Notice that it suffices to show that there will always be $>1$ black triangle on either side of $t$ since otherwise we could choose a large enough triangle $\mathcal{S}$ covering all but this alone triangle, where one side of $\mathcal{S}$ will lie on $t$ . We could have applied the same argument to lines possessing, let's say $r$, property which are horizontal lines basically, but notice that this is redundant because either shifting lines possessing property $r$ up or down will increase/decrease the number of triangles above or below these lines by $(\tfrac{n}{2})$ due to our nice configuration, not something we wanted, keeping this aside, now we proceed with this multi-step process using our shifting algorithm : Take $t$ from either left side of leftmost pair (the one described earlier) or right side of right most pair and began shifting it right or left according to the need, basically we are exhausting lines for triangle $\mathcal{S}$. For, if $t$ is from left side of leftmost pair, notice that shifting it right will always increase/decrease an even number of black triangles on either side of $t$ (due to our construction). So if we take $2$ vertices of $\mathcal{S}$ on $t$ and the third one on either side of $t$ we will always miss an even number of black triangles on one side of $t$ and outside of $\mathcal{S}$ forcing $n$ to be ngood $~~$ For, if $t$ is from right side of rightmost pair, same argument as above works here forcing $n$ to be ngood Notice that this works for $n\ge 4 ~(\text{and even obviously, due to assumption})$ because of shifting algorithm $\square$ We will follow the similar argument for $n$ odd. Claim : For $n\ge5$ and odd, is ngood Proof : Here, first of all, place $(\tfrac{n-1}{2})$ unit black triangles desribed earlier both above and below of $\ell$, and place $1$ extra unit black triangle either above or below of $\ell$ and somewhere between those $(\tfrac{n-1}{2})$ unit black triangles recently chosen. Call this extra unit black triangle special. It's clear that $\exists$ atleast one nice pair either side of this due to $n\ge 5$. Now applying the shifting algorithm, as earlier, clearly we will face no problem (in the sense that there doesn't exists any $\mathcal{S}~$) until we reach special triangle from either sides but even this is not any problem since either side we will have $\ge 2$ black triangles contrary to what we want, that is, one unit black triangle on one side $\square$ We just need to check $n\le 3$. For $n=1$, is trivially true. For $n=2$ we could choose one of the two unit black triangles, hence true. For $n=3$, let $r$ be any line in any of $\mathbb{P},\mathbb{Q},\mathbb{R}$. Here $\mathbb{R}$ is the set of all horizontal lines (recall that we have tiled the grid according to our need). Let $r'$ be any another line in the same set. Call the two lines $\{r,r'\}$, along with the space between them, slide iff the $r$ and $r'$ are consecutive. Now if all $3$ unit triangles belongs to different slides, choice of $\mathcal{S}$ is rather obvious. If $2$ of $3$ triangles are in the same slide, then choose $\mathcal{S}$ such that it contains those $2$ triangles and it's third vertex lies on the opposite side of the third triangle with respect to the slide conatining $2$ triangles. If all $3$ unit black triangles lie in same slide, again choose $\mathcal{S}$ such that it covers either first and second but not third, or second and third but not first, exhausting all cases. We have, as solution, $n=1,2,3$ only works $\blacksquare$