A party has $2 n$ people attending. If any two people at the party have the same number of friends, then these two people do not have any common friends (friendship is mutual). How many pairs of friends can there be at most in this party?
(Wu Yunjian, contributor)
The maximum number of friend pairs sought is $\frac{n(n+1)}{2}$.
We use $2n$ points to represent $2n$ people. If two people are friends, we draw an edge between their corresponding points, and call one point a neighbor of the other. This way, we get a simple graph $G$ with these $2n$ people as the vertex set, satisfying the following condition: if two people have the same number of neighbors, then these two people's neighbor sets have no common elements.

Assume $A$ is the point with the most neighbors in the graph. Let $A$ have $k$ neighbors. We will prove that $k \leqslant n$, and that these $k$ points have neighbor counts of $1, 2, \cdots, k$.

Let the $k$ neighbors of $A$ be $A_{1}, A_{2}, \cdots, A_{k}$. Since $A_{1}, A_{2}, \cdots, A_{k}$ have a common neighbor $A$, the number of neighbors of these points are all different. It is also easy to see that the number of neighbors of these points is at least 1 (all have neighbor $A$) and at most $k$ (since $A$ has the most neighbors), so the number of neighbors of $A_{1}, A_{2}, \cdots, A_{k}$ is a permutation of $1, 2, \cdots, k$.

Without loss of generality, assume $A_{i}$ has exactly $i$ neighbors $(i=1,2, \cdots, k)$. Consider $A_{k}$. At this point, $A_{k}$ has the same number of neighbors as $A$, so they have no common neighbors, meaning $A_{k}$ has no neighbors in $\left\{A_{1}, A_{2}, \cdots, A_{k-1}\right\}$. Therefore, the total number of people
$$
2n \geqslant 1 + k + (k-1) = 2k,
$$

hence $k \leqslant n$.
Let the total number of neighbor pairs be $S$. From the above, $A$'s neighbors $A_{1}, A_{2}, \cdots, A_{k}$ have $1, 2, \cdots, k$ neighbors respectively, corresponding to $1, 2, \cdots, k$ neighbor pairs. Let the $k-1$ neighbors of $A_{k}$ other than $A$ be $B_{1}, B_{2}, \cdots, B_{k-1}$.

For the $2n-2k$ points other than $A, A_{1}, A_{2}, \cdots, A_{k}, B_{1}, B_{2}, \cdots, B_{k-1}$, each point has at most $k$ neighbors, and can produce at most $\frac{k(2n-2k)}{2}$ neighbor pairs. Therefore,
$$
\begin{aligned}
S & \leqslant 1 + 2 + \cdots + k + \frac{k(2n-2k)}{2} = \frac{k(2n-k+1)}{2} \\
& \leqslant \frac{(k+2n-k+1)^2}{8} = \frac{(2n+1)^2}{8} = \frac{4n(n+1)+1}{8}.
\end{aligned}
$$

Since $8 \nmid 4n(n+1)+1$, we have
$$
S \leqslant \left[\frac{4n(n+1)+1}{8}\right] = \frac{n(n+1)}{2}.
$$

Finally, we show that $S$ can achieve $\frac{n(n+1)}{2}$.
Consider the following simple graph with $2n$ points $A_{1}, A_{2}, \cdots, A_{n}, B_{1}, B_{2}, \cdots, B_{n}$ as vertices. When $i \geqslant \left[\frac{k}{2}\right]+1$, the neighbor set of $A_{i}$ is $\left\{B_{i}, A_{n-i+1}, A_{n-i+2}, \cdots, A_{n}\right\} \backslash \left\{A_{i}\right\}$, and the neighbor set of $B_{i}$ is $\left\{A_{i}, B_{n-i+1}, B_{n-i+2}, \cdots, B_{n}\right\} \backslash \left\{B_{i}\right\}$; when $i \leqslant \left[\frac{k}{2}\right]$, the neighbor set of $A_{i}$ is $\left\{A_{n}, A_{n-1}, \cdots, A_{n-i+1}\right\}$, and the neighbor set of $B_{i}$ is $\left\{B_{n}, B_{n-1}, \cdots, B_{n-i+1}\right\}$. At this point, $A_{i}$ and $B_{i}$ each have $i$ neighbors $(i=1,2, \cdots, n)$, and the total number of neighbor pairs $S = \frac{1}{2} \times 2(1+2+\cdots+n) = \frac{n(n+1)}{2}$.
In conclusion, the maximum number of friend pairs sought is $\frac{n(n+1)}{2}$.