For a positive integer $n$, there is a school with $2n$ people. For a set $X$ of students in this school, if any two students in $X$ know each other, we call $X$ [i]well-formed[/i]. If the maximum number of students in a well-formed set is no more than $n$, find the maximum number of well-formed set. 

Here, an empty set and a set with one student is regarded as well-formed as well.
1. **Graph Representation and Definitions**:
   - Let $\mathcal{G}$ be the graph representing the school where vertices represent students and edges represent the acquaintance between students.
   - A set $X$ of students is called *well-formed* if any two students in $X$ know each other, i.e., $X$ forms a clique in $\mathcal{G}$.
   - The problem states that the maximum size of a well-formed set (clique) is no more than $n$.

2. **Complement Graph and Matching**:
   - Consider the complement graph $\overline{\mathcal{G}}$ of $\mathcal{G}$, where an edge exists between two vertices if and only if there is no edge between them in $\mathcal{G}$.
   - We claim that the maximum number of well-formed sets is $3^n$. This can be achieved by choosing the complement of $\mathcal{G}$ to be a matching.

3. **Vertex Clowning**:
   - Define a *clown* of a vertex $u$ to be a vertex having the same neighborhood as $u$ but not adjacent to $u$.
   - A vertex $x$ is *stable* if in the complementary graph $\overline{\mathcal{G}}$, $x$ belongs to some clique that is disconnected from the rest of the graph.

4. **Clowning Process**:
   - Let $f(u)$ denote the number of cliques containing $u$.
   - Choose $u \in \arg\max_{x \text{ not stable}} f(x)$ and in each step, choose $v \not \sim u$ such that $v$ is not a clown of $u$, then replace $v$ by a clown of $u$.
   - Draw a red edge between pairs of clowns of $u$ that have not received a red edge between them unless there is some new vertex $x$ not stable with $f(x) > f(u) = f(v)$ caused by the clowning, in which case erase all red edges and restart the process with $x$ instead of $u$.

5. **Properties and Termination**:
   - There are no $n+1$-cliques: If a new $n+1$-clique is created, it can use exactly one of $u, v$. This leads to a contradiction.
   - The number of cliques in the graph does not increase because in the clowning process, all $f(v)$ cliques originally containing $v$ were removed, and then $f(u) \ge f(v)$ containing $v$ were created.
   - If $v_1v_2, v_1v_3$ are both red edges, then so is $v_2v_3$.

6. **Algebraic Analysis**:
   - The complementary graph of the resulting $\mathcal{G}$ is a union of pairwise disconnected cliques of sizes $n_1 \ge \dots \ge n_k > \underbrace{1, \dots, 1}_{s \text{ copies}}$.
   - The number of cliques in $\mathcal{G}$ is:
     \[
     2^s \prod_{i=1}^k (n_i + 1) \le 2^s \cdot \frac{(\sum n_i + k)^k}{k^k} = \frac{2^s(2n - s + k)^k}{k^k}
     \]
     with $\sum_{i=1}^k n_i = 2n - s$ and $k + s \le n$.
   - We show that $s = n - k$ when the maximum of the RHS is reached. For $k + s \le n - 1$:
     \[
     \frac{2^{s+1}(2n - s - 1 + k)^k}{2^s(2n - s + k)^k} > 1
     \]
     follows from:
     \[
     \frac{(2n - s + k)^k}{(2n - s - 1 + k)^k} = \left(1 + \frac{1}{2n - s - 1 + k}\right)^k \le \left(1 + \frac{1}{2k}\right)^k < \sqrt{e} < 2
     \]

7. **Maximization**:
   - It suffices to maximize:
     \[
     \frac{2^n (n + 2k)^k}{(2k)^k}
     \]
     Since the function $\left(1 + \frac{a}{x}\right)^x$ is increasing for any $a > 1$, the maximum of the above as a function of $k$ is reached when $k = n$, whence the quantity takes the value of $3^n$.

Thus, the maximum number of well-formed sets is $3^n$.

The final answer is $\boxed{3^n}$