Show that there are at most $n^2$ roads.

Let us form a graph $V$ where $V= \{1,2,\ldots,2n\}$, and $i\leftrightarrow j$ if $i$-th place and $j$-th place are connected by a road between them. Each place is not connected with itself. Let $d_i$ denote the number of $j$ such that $i\leftrightarrow j$. Then we can see that the number of edges is $$N = \frac{1}{2}\sum\limits_{i\in V}d_i.$$ Our goal is to show $N\le n^2$. Let us define $$ M=\sum_{i\leftrightarrow j}\sum_{k\in V}\left(1_{i\leftrightarrow k}+1_{j\leftrightarrow k}\right) $$ where $1_{j\leftrightarrow k}=1$ if $j\leftrightarrow k$ and is $0$ otherwise. By the assumption, there is no $k$ such that $ 1_{i\leftrightarrow k}+1_{j\leftrightarrow k}=2.$ This gives $$ M\le \sum_{i\leftrightarrow j}\sum_{k\in V}1=2n\sum_{i\in V} d_i= 4nN. $$ On the other hand, we have $$\begin{eqnarray} M=\sum_{k\in V}\sum_{i\leftrightarrow j}\left(1_{i\leftrightarrow k}+1_{j\leftrightarrow k}\right)&=&\sum_{k\in V}\sum_{i\leftrightarrow j}1_{i\leftrightarrow k} +\sum_{k\in V}\sum_{i\leftrightarrow j}1_{j\leftrightarrow k}\\ &=&2\sum_{k\in V}\sum_{i\leftrightarrow j}1_{i\leftrightarrow k}\\ &=&2\sum_{k\in V}\sum_{i\in V}1_{i\leftrightarrow k}\sum_{j:i\leftrightarrow j}1\\ &=&2\sum_{k\in V}\sum_{i\in V}1_{i\leftrightarrow k}\cdot d_i\\ &=&2\sum_{i\in V}d_i\sum_{k\in V}1_{i\leftrightarrow k}\\ &=&2\sum_{i\in V}d_i\cdot d_i = 2\sum_{i\in V} d_i^2. \end{eqnarray}$$ By Cauchy-Schwarz, we have $$ n\cdot M=|V|\cdot\sum_{i\in V} d_i^2\ge \left(\sum_{i\in V}d_i\right)^2=4N^2 $$ Since $M\le 4nN$, it follows $$ 4N^2\le 4n^2N, $$ that is, $$ N\le n^2 $$as desired.

It's equivalent to prove that in graph with $2n$ vertices without triangles there at most $n^2$ edges. However, it's a case of Turán's theorem (about $K_{r}$-free graphs) for $r=3$. You can find several proofs of this theorem in Proofs from THE BOOK by Aigner and Ziegler.

https://en.m.wikipedia.org/wiki/Turán%27s_theorem