Can a subset be chosen that intersects with other subsets a given number of times? Let $S$ be a set，$|S| = n$, with $n$ sufficiently large and divisible by $8$. Suppose that $A_1, \dots, A_{n/2} \subseteq S$, $|A_i| = \dfrac{n}{2}$ for all $i$. Is it always possible to choose a subset $T \subseteq S$ with $|T| = \dfrac{n}{4}$ such that $|T \cap A_i| \geq \dfrac{n}{8}$ for all $i$?
I'm considering a set of integers from $1$ to $n$. If each $A_i = \{i, i+1, \dots, i + n/2\}$, then choosing $T$ to be every 4th integer would do the trick; that is, we can let $T = \{1,5,9,13,\dots\}$. It would also work to let $T$ be spread more coarsely across the integers, for example letting $T = \{1,2,9,10,17,18,\dots\}$, or even letting $T = \{1, \dots, n/8, n/2 + 1, \dots, 5n/8\}$. I'm trying to see if this idea is true in general.
I'm having problems using a standard counting argument, because as I build $T$, I have to keep track of which $A_i$'s already intersect with $T$ a sufficient number of times and don't need to be considered anymore. Furthermore, I know that the number of sets, $n/2$, is critical to the argument, because this would definitely not work if there were, say, $2^n$ subsets of $S$.
If there are any known results related to this kind of idea, or if anyone has any clever arguments, I would be extremely grateful to hear them. Thank you.
 A: No, for sufficiently large $n$ there is always some choice of $A_1, \dots, A_{n/2}$ for which no such $T$ exists.
Consider the case where $A_{n/4+i} = S \setminus A_i$ for $1 \leq i \leq n/4$, so here the question is whether there is some $T \subset S$ of size $n/4$ for which $|T \cap A_i| = n/8$ (note the '$=$' sign) for each $1 \leq i \leq n/4$.
Let $A_1, \dots, A_{n/4}$ be chosen uniformly and independently at random from all subsets of $S$ of size $n/2$ (there are $\binom{n}{n/2}$ possible sets). For now, fix a subset $T \subset S$ with $|T| = n/4$, and let $E_T$ be the event that $|T \cap A_i| = n/8$ for $1 \leq i \leq n/4$. Then using Stirling's approximation, which gives $\binom{2n}{n} \sim \frac{4^n}{\sqrt{\pi n}}$, we have 
$$\mathbb{P}[|T \cap A_i| = n/8] = \frac{\binom{n/4}{n/8} \binom{3n/4}{3n/8}}{\binom{n}{n/2}} \sim \frac{1}{\sqrt{\frac{3}{32} \pi n}}$$
so in particular there is some $c > 0$ with $\mathbb{P}[|T \cap A_i| = n/8] \leq \frac{c}{\sqrt{n}}$. But by independence of $A_1, \dots, A_{n/4}$, 
$$\mathbb{P}[E_T] = \prod_{i=1}^{n/4} \mathbb{P}[|T \cap A_i| = n/8] \leq \left(\frac{c}{\sqrt{n}}\right)^{n/4}$$
and thus the probability that $E_T$ holds for some $T$ is bounded by 
$$\mathbb{P}[\text{$E_T$ for some $T$}] \leq \sum_{T} \mathbb{P}[E_T] \leq \binom{n}{n/4}\left(\frac{c}{\sqrt{n}}\right)^{n/4} \leq 2^n \left(\frac{c}{\sqrt{n}}\right)^{n/4} = 2^{-(n/8)\log n + O(n)} < 1$$
for sufficiently large $n$. It follows that there are some $A_1, \dots, A_{n/4}$ for which no $E_T$ holds, i.e. there is no $T$ of size $n/4$ with all $|T \cap A_i| = n/8$. 
Note: The same proof also shows the stronger result that if $k$ is such that for any $A_1, \dots, A_k \subset S$ of size $n/2$, there is some $T$ of size $n/4$ with all $|T \cap A_i| \geq n/8$, then $k = O(n/\log n)$. I would guess that this bound can be pushed lower.
