This looks quite hard. Does anyone have any thoughts or suggestions?

Nice problem The answer is $((2^m+1)^{k-1})(2^{m-1})$ For construction, fix a blue element of $S_1$. Notice that we have $2^{m-1}$ choices for $S_1$ and $2^m+1$ choices for each of $S_i$ with $i>1$. For proving the bound, I will induct on $k$. Base $k=2$: Let $S_1={a,a_1,a_2,\ldots,a_m}$ and $S_2={b,b_1,b_2,\ldots,b_m}$ with $a,b$ being red. If we have both $a$ and $b$ in the union of elements of $\mathcal{G}$, we clearly have a set $T\in\mathcal{G}$ such that $a,b\in T$. Then we get that each element of $\mathcal{G}$ has one of $a$ or $b$. WLOG let them all have $b$. So the maximum number of elements of $\mathcal{G}$ will be $2^m+1$. If we don't have both $a$ and $b$ in the union, one can easily check that we have at most $2^{m-1}(2^m+1)$ elements in $\mathcal{G}$ since in the best case, each two elements of $\mathcal{G}$ should intersect in $S_1$ (I have assumed that we don't have $a$ in the union) and so we have at most $2^{m-1}$ choices for $S_1$ and $2^m+1$ choices for $S_2$. And thus the base $k=2$ is done. In order to complete the induction, Note that the number of elements $P\in\mathcal{G}$ such that $P\cap S_1=a$ is at most $f(k,m)$. Because if it is more than $f(k,m)$ then two of this elements have only $a$ as their intersection and thus each other two element $Q,R\in\mathcal{G}$ that do not contain $a$ have intersection of size at least $2$ among $S_2,\ldots,S_{k+1}$. I think by counting the maximum possible number of this subsets, you see that the total number is less than $((2^m+1)^{k-1})(2^{m-1})$ and thus it is not the best case so that number is at most $f(k,m)$. Now pair each set of $a_1,a_2,\ldots,a_m$ like $T$ with $S_1-a-T$. Note that this two do not intersect and for each two non-intersecting sets like $K$ and $L$ we have: $$\text{The number of sets in } \mathcal{G}\text{ having intersection }K\text{ with } S_{k+1}+\text{ The number of sets in } \mathcal{G}\text{ having intersection }L\text{ with } S_{k+1}\le 2f(k,m)$$. In other words we take a complete matching among the $2^m$ sets, and each pair, has at most $2f(k,m)$ elements. By summing up, you see that we have at most $2^{m-1}(2f(k,m))+f(k,m)=(2^m+1)f(k,m)$ elements and we are done by the induction hypothesis.

ayeen_izady wrote: Lemma: The $2^m+1$ monochromatic subsets of $S_1$ can be written around a circle such that for each two neighbors like $K$ and $L$ we have $$K\cap L=\emptyset$$Proof: You can prove the lemma by inducting on $m$. Can you provide this construction explicitly for the case of $m = 3$?

YaoAOPS wrote: ayeen_izady wrote: Lemma: The $2^m+1$ monochromatic subsets of $S_1$ can be written around a circle such that for each two neighbors like $K$ and $L$ we have $$K\cap L=\emptyset$$Proof: You can prove the lemma by inducting on $m$. Can you provide this construction explicitly for the case of $m = 3$? Is it fixed now?

The answer is $\frac{2^{m-1}}{2^m+1} (2^m+1)^k$. The construction is simply picking all the sets with the same blue element. Now we prove that this is the upper bound. Suppose that we pick a proportion $p$ of all the sets in $\mathcal{F}$; it suffices to prove that $p \le \frac{2^{m-1}}{2^m+1}$. For a set $S\in \mathcal{F}$, let the indicator function $f(S)$ be $1$ if $S\in \mathcal{G}$ and $0$ otherwise. We define random variables $X_i(1\le i \le 2^m+1)$ by picking sets $Y_i\in \mathcal{F}$ according to some distribution defined later and let $X_i=f(Y_i)$. For $1\le i \le k$, let $Z_i=(Y_1\bigcap S_i,Y_2\bigcap S_i,\cdots,Y_{2^m+1}\bigcap S_i)$ be independent random variables, and $Z_i$ is sampled as follows. Let $R_i$ be the set formed by the unique element in $S_i$ and $B_i$ be those blue elements. Pick a uniformly random $x\in [2^m+1]$ and a uniformly random $Y\subseteq B_i$ and let $Z_i$ be $(R_1,Y,B_i\setminus Y,Y,B_i\setminus Y,\cdots,Y,B_i\setminus Y)$(repeating $2^{m-1}$ times) cyclic shifted by $x$. We can easily see that $Y_i$ is the uniform distribution on $\mathcal{F}$ for each $1\le i\le 2^m+1$ since each coordinate of $Z_i(1\le i \le k)$ is the uniform distribution on the monochromatic subsets of $S_i$, so $\mathbb E(X_i)=p$. However, $\sum\limits_{i=1}^{2^m+1}X_i \le 2^{m-1}$ always holds as we can never have $Y_i,Y_{i+1}\in \mathcal{G}$ simultaneously. So we conclude by the linearity of expectation.