3. Given a positive integer $n$. For a family of subsets $\mathscr{D}$ of the set $\{1,2, \cdots, n\}$, if it satisfies the following condition: if $A$ is a member of the family $\mathscr{D}$, and $B$ is a subset of $A$, then $B$ is also a member of the family $\mathscr{D}$, then the family $\mathscr{D}$ is called "downward closed". For a downward closed family $\mathscr{D}$, find the maximum value that the expression $\sum_{A \in \mathscr{D}}(-1)^{|A|}$ can take. Here, $|A|$ denotes the sum over all members $A$ of the family $\mathscr{D}$.

Note: A family of subsets of a set $X$ refers to a set consisting of several subsets of $X$.
The problem is from the 2015 Tsinghua University Mathematics Autumn Camp. Since $\mathscr{D}$ is any downward closed subfamily, let $f(k)$ denote $\max _{A \in \mathscr{D}}|A|=k$ when $\sum_{A \in \mathscr{E}}(-1)^{|A|}$, then we need to find $\max _{0 \leqslant k \leqslant n} f(k)$.

By classifying $k$ based on its parity, it is easy to see that when $k$ is odd, $f(k)$ cannot achieve the maximum value $\max _{0 \leqslant k \leqslant n} f(k)$.

Subsequently, we can prove by mathematical induction that when $k$ is even and $k \leqslant n$,
$$
f(k)=\sum_{i=0}^{k}(-1)^{i} \mathrm{C}_{n}^{i} \quad (0 \leqslant k \leqslant n).
$$

Let $\{1,2, \cdots, n\}$ be the family of all subsets with at most $k$ elements, denoted as $\mathscr{D}_{k}$. When $\mathscr{D}=\mathscr{D}_{k}$, $\sum_{A \in \mathscr{\mathscr { N }}}(-1)^{|A|}$ reaches its maximum value $f(k)$.
Thus, we have
$$
\max _{0 \leqslant k \leqslant n} f(k)=f\left(2\left[\frac{n}{4}\right]\right)=\sum_{i=0}^{2\left[\frac{n}{4}\right]}(-1)^{i} \mathrm{C}_{n}^{i},
$$

where $[x]$ denotes the greatest integer not exceeding the real number $x$.