Problem 1071

4. Let $A=\{1,2, \cdots, 4 n\}$. Let $\mathcal{F}$ be the family of subsets in $A$ with $n+1$-elements. Then
$$|\mathcal{F}|=\binom{4 n}{n+1} .$$

Note that every $n+1$-element in $S_{i}$ is also a member in $\mathcal{F}$. Since $S_{i} \cap S_{j}$ contains at most $n$ elements in $A$, and any $n+1$-element in $S_{i}$ is different from any $n+1$-element in $S_{j}$ for all $1 \leq i<j \leq k$. Thus
$$|\mathcal{F}| \geq \sum_{i=1}^{k}\binom{2 n}{n+1}=k\binom{2 n}{n+1} .$$

Hence
$$\begin{aligned} k & \leq\binom{ 4 n}{n+1} \div\binom{ 2 n}{n+1} \\ & -\frac{4 n \times(4 n-1) \times \cdots \times 3 n}{2 n \times(2 n-1) \times \cdots \times n} . \end{aligned}$$

It can be shown that
$$\frac{(4 n-i)(3 n+i)}{(2 n-i)(n+i)} \leq \frac{4 n \times 3 n}{2 n \times n}=6$$
for all $0 \leq i \leq(n-1) / 2$.
If $n$ is odd, then
$$\frac{4 n \times(4 n-1) \times \cdots \times 3 n}{2 n \times(2 n-1) \times \cdots \times n}=\prod_{i=0}^{(n-1) / 2} \frac{(4 n-i)(3 n+i)}{(2 n-i)(n+i)} \leq 6^{(n+1) / 2} .$$

If $n$ is even, then
$$\frac{4 n \times(4 n-1) \times \cdots \times 3 n}{2 n \times(2 n-1) \times \cdots \times n}=\frac{4 n-n / 2}{2 n-n / 2} \prod_{i=0}^{(n-2) / 2} \frac{(4 n-i)(3 n+i)}{(2 n-i)(n+i)} \leq 6^{1 / 2} 6^{n / 2}=6^{(n+1) / 2}$$