Diagonal-intersection set-pair bound ($|I|\le n-r-s+2$)

Let $I$ index at least two pairs of subsets of an $n$-element ground set, with $|R_i|=r$, $|S_i|=s$, and $R_i\cap S_j\ne \varnothing$ exactly when $i=j$. ThenChoose a witness $x_i\in R_i\cap S_i$ for each pair. The intersection conditions make these witnesses distinct and exclude $x_i$ from both sets belonging to every other index. For distinct $i,j$, the sets $R_i\setminus \{x_i\}$ and $S_j\setminus \{x_j\}$ are disjoint and both lie outside the entire witness set. Consequently $n-|I|\ge (r-1)+(s-1)$.

For sharpness, choose disjoint fixed sets $P,Q$ of sizes $r-1,s-1$ and let the remaining points be witnesses. Put $R_i=P\cup \{x_i\}$ and $S_i=Q\cup \{x_i\}$. This construction attains equality whenever there are at least two witnesses. The two-pair hypothesis cannot be dropped: a single pair $R_1=S_1=[n]$ satisfies the intersection condition, but for $n\ge 2$ the proposed unrestricted bound would read $1\le 2-n$.