$\newcommand{\lrp}[1]{\left(#1\right)}$ $\newcommand{\R}{\mathbf R}$ $\newcommand{\lrb}[1]{\left[#1\right]}$
Problem (Iberoamerican Olympiad 2001). Let $n, r, k$ be positive integers and assume $k\geq 2$. Let $S$ be a set with $n$ elements and $S_1, \ldots, S_k$ be subsets of $S$, each of size $r$. Show that there exists $i$ and $j$ such that $1\leq i<j\leq k$ satisfying $$ |S_i\cap S_j| \geq r - \frac{nk}{4(k-1)} $$
What I tried gives a bound but does not necessarily achieve the desired result.
Let $1_{S_i}:S\to \R$ be the indicator function of $S_i$. We have $$ \lrp{\sum_{i=1}^k 1_{S_i}}^2 = \sum_{i=1}^k 1_{S_i} + 2\lrp{\sum_{1\leq i<j\leq k} 1_{S_i\cap S_j}} $$ Operating $\sum_{x\in S}$ on both sides, and writing $\sum_{i=1}^k 1_{S_i}(x)$ as $d(x)$, we get $$ \sum_{x\in S} d(x)^2 = rk + 2\lrp{\sum_{1\leq i< j \leq k} |S_i\cap S_j|} $$ Using $\sum_{x\in S}d(x)^2 \geq \frac{1}{|S|}\lrp{\sum_{x\in S} d(x)}^2 = r^2k^2/n$, we have $$ \frac{r^2k^2}{n} \leq rk + 2\lrp{\sum_{1\leq i<j\leq k} |S_i\cap S_j|} $$ Therefore $$ \frac{1}{\binom{k}{2}}\sum_{1\leq i<j\leq k} |S_i\cap S_j| \geq \frac{1}{k(k-1)} \lrb{ \frac{r^2k^2}{n} - rk} $$ and hence there must exists $1\leq i<j\leq k$ such that $$ |S_i\cap S_j| \geq \frac{1}{k(k-1)} \lrb{ \frac{r^2k^2}{n} - rk} $$ Upon making some calculation, it follows that this bound is better than the desired bound if $r\geq n/2$ and worse otherwise.
EDIT: I made a mistake in my last line. The bound I achieve is actually better than the desired bound always.
The following simple computation proves the last claim in the question, "the bound I achieve is actually better than the desired bound always.". $$\begin{aligned} &\quad\frac{1}{k(k-1)}\left(\frac{r^2k^2}{n} - rk\right) -\left(r - \frac{nk}{4(k-1)}\right) \\ &=\frac{1}{4n(k-1)}\left(4r^2k-4nr\right)-\frac1{4n(k-1)}\left(4n(k-1)r-n^2k\right) \\ &=\frac{k}{4n(k-1)}(2r-n)^2 \\ &\ge0. \end{aligned}$$
The equality $\displaystyle{\max_{i,j}|S_i\cap S_j|=r-\frac{nk}{4(k-1)}}$ can hold, for example, when $S=\{1,2,3,4,5,6\}$, $S_1=\{1,2,3\}$, $S_2=\{3,4,5\}$, $S_3=\{5,6,1\}$, $S_4=\{2,4,6\}$.
All other computations in the question are valid, too.