A) Let $E$ be the family of sets $a_{i}= \{1, 2, \cdots, i\}, i = 1, 2, \cdots, n$. Given any set $S$ with $|S| = k$. Then $|S \bigtriangleup a_{1}| = k \pm 1, |S \bigtriangleup a_{i+1}| = |S \bigtriangleup a_{i}| \pm 1$, and $|S \bigtriangleup a_{n}| = n-k$. So either one of either $|S \bigtriangleup a_{1}|$ or $|S \bigtriangleup a_{n}|$ is equal to $\frac{n}{2}$, or they lie on opposite sides of $\frac{n}{2}$. In the latter case, one of the intermediate $a_{i}$ must be $\frac{n}{2}$. B) Let $E$ be the family of sets ${a_{i}= \{1, 2, \cdots,1+(d-1) i}$ such that $0 \leq i < \lceil \frac{n}{d+1}\rceil = j$. Given any set $S$ with $|S| = k$. Then $|S \bigtriangleup a_{0}| = k \pm 1, | |S \bigtriangleup a_{i+1}|-|S \bigtriangleup a_{i}| | \leq d+1$, and $| |S \bigtriangleup a_{j}|-(n-k) | \leq d$. Then both $|S \bigtriangleup a_{0}|$ and $|S \bigtriangleup a_{0}|$ cannot be less than $\frac{n-d}{2}$, since their sum is at least $n-d-1$. Similarly, they cannot both be more than $\frac{n-d}{2}$. So either one of them lies within the given range, or they lie on opposite sides. So one of the intermediate $a_{i}$ must lie within the range, as $d+1$ is not long enough to cross it.