3. Let $E$ be a given $n$-element set, and $A_{1}, A_{2}, \cdots, A_{k}$ be $k$ distinct non-empty subsets of $E$, satisfying: for any $1 \leqslant i<j \leqslant k$, either $A_{i} \cap A_{j}=\varnothing$ or one of $A_{i}$ and $A_{j}$ is a subset of the other. Find the maximum value of $k$.
(Cold Gangsong, problem contributor)
3. The maximum value of $k$ is $2 n-1$.

Let $E=\{1,2, \cdots, n\}$.
For example, when $1 \leqslant i \leqslant n$, let $A_{i}=\{i\}$;
when $n+1 \leqslant i \leqslant 2 n-1$, let $A_{i}=\{x \in E \mid 1 \leqslant x \leqslant i-n+1\}$.
The $2 n-1$ sets $A_{i}$ thus selected satisfy the conditions.
We prove by mathematical induction that $k \leqslant 2 n-1$.
When $n=1$, the conclusion is obvious.
Assume that when $n \leqslant m-1$, the conclusion holds.
When $n=m$, consider the set among $A_{1}, A_{2}, \cdots, A_{k}$ that is not the full set and has the most elements (denote it as $A_{1}$, and assume $A_{1}$ has $t$ elements). Then $t \leqslant m-1$.
We classify $A_{1}, A_{2}, \cdots, A_{k}$ into three categories:
(1) The full set;
(2) Not the full set, and the intersection with $A_{1}$ is not empty;
(3) The intersection with $A_{1}$ is empty.
By the selection of $A_{1}$, we know that the sets in (2) are all subsets of $A_{1}$ and still satisfy the conditions. By the induction hypothesis, the number of sets in (2) does not exceed $2 t-1$.

The sets in (3) are all subsets of $E \backslash A_{1}$. By the induction hypothesis, the number of sets does not exceed $2(m-t)-1$.
$$
\begin{array}{l}
\text { Hence } k \leqslant 1+(2 t-1)+[2(m-t)-1] \\
=2 m-1,
\end{array}
$$

That is, when $n=m$, the conclusion also holds.
Therefore, for any $n$-element set $E$, $k \leqslant 2 n-1$.