What is the maximum number of non-empty subsets that can be selected from a 100-element set such that any two selected subsets are either disjoint or one contains the other?
Let the 100-element set be the set of numbers $1,2, \ldots, 100$. Select all its one-element subsets, as well as the sets of the form $\{1,2, \ldots, n\}$, where $n=2,3, \ldots, 100$. The 199 subsets thus selected satisfy the conditions of the problem. We will show that it is not possible to select 200 subsets that satisfy the conditions of the problem, so the answer to the problem is 199.

In general, we will show that from an $n$-element set, at most $2n-1$ subsets can be selected that meet the conditions of the problem. We prove this by induction, for $n=1$ this is clearly true. Now suppose that the statement is true for all natural numbers less than $n$, and select as many subsets as possible from an $n$-element set $H$ such that the subsets satisfy the conditions. It is obvious that $H$ will be among the selected ones, since if it were not, we could add it, as every selected subset is a subset of $H$. We will prove that at most $2n-2$ proper subsets can appear. Let $A$ be one of the largest among the selected proper subsets, and denote the number of elements in $A$ by $a$. According to the condition, every other subset is either a subset of $A$, or contains $A$, or is disjoint from it. But a proper subset containing $A$ cannot exist, since the number of elements in $A$ was maximal. Therefore, among the other selected subsets, there are two types: those that are subsets of $A$, and those that are disjoint from it.

Those that are subsets of $A$, together with $A$, form a subsystem of $A$ that satisfies the conditions of the problem. The number of elements in $A$ is less than $n$, so by the induction hypothesis, there can be at most $2a-1$ sets in such a subsystem. The subsets that are disjoint from $A$ also satisfy the conditions of the problem, and are all subsets of the complement of $A$, $H-A$, which has $n-a \geq 1$ elements, since $A$ is not empty. Thus, from $(H-A)$, by the induction hypothesis, at most $2(n-a)-1$ subsets can be selected that meet the requirements of the problem. We have found that there can be at most $2a-1$ subsets that are subsets of $A$, and at most $2(n-a)-1$ that are disjoint from $A$. Together with $H$, therefore, we can select at most $1+(2a-1)+2(n-a)-1=2n-1$ proper subsets, as we claimed.

This proves that from an $n$-element set, at most $2n-1$ non-empty subsets can be selected that meet the requirements of the problem.

Remarks. 1. The example given at the beginning of the solution can obviously be generalized: from an $n$-element set, $2n-1$ subsets can always be selected that meet the requirements of the problem.

2. By thinking through the proof, we can generally specify all $2n-1$-element subsystems of subsets that satisfy the requirements of the problem: we select the entire set, and at each step, we arbitrarily divide a selected but not yet "cut" set into two parts, and select both of the resulting subsets. We continue this process until every "uncut" set becomes a one-element set. This way, as can be easily seen, we get $2n-1$ subsets, which is good, and we cannot get $2n-1$ good subsets any other way.
3. During the proof, we "helped ourselves" twice with a general idea. First, we used complete induction, and second, we cleverly chose the set $A$: we said it should be the largest possible. With this, we made the system of sets more comprehensible to us. Both ideas are often useful in solving combinatorial problems.