I believe this is the same as a problem from Ukraine TST 2015. However, this thread is clearly better

The answer is $2\left\lfloor \dfrac{n}{2}\right\rfloor + 1$. If $n$ is even, we divide $2n$ teams into $2$ groups $\left\{A_1,A_2,\ldots,A_{n-1}\right\}$ and $\left\{B_1,B_2,\ldots,B_{n+1}\right\}$. In the $i$th round, schedule the matches: $\left\{A_1, B_i\right\},\ldots, \left\{A_i, B_{i+ n-2}\right\},\left\{B_{i+n-1}, B_{i+n}\right\}$. After $n+1$ rounds, the pairs between unplayed teams form a graph which is the union of a $\left(n-1\right)$-clique and $\left(n+1\right)$-clique. Note that both of these two cliques do not have a perfect matching, thus, no more rounds can be scheduled. To prove $n+1$ optimal, we need the following lemma: Given a $\left(n-1\right)$-regular graph with $2n$ vertices. Suppose that the graph doesn't have a perfect matching. Then the graph is the union of two $n$-clique. Observe that if $n$ is even, the $n$-clique itself has a perfect matching, hence, after any $2n$ rounds, the graph forming by unplayed pairs is $\left(2n-1\right)$-regular, so it still contain a perfect matching and we can schedule one more round. The method using for $n$ odd is similar.

internationalnick123456 wrote: Given a $\left(n-1\right)$-regular graph with $2n$ vertices. Suppose that the graph doesn't have a perfect matching. Then the graph is the union of two $n$-clique. Where can I find this lemma's proof?

I don't know the source, however, you can use Exercise 4 in here to prove it.

nAalniaOMliO wrote: internationalnick123456 wrote: Given a $\left(n-1\right)$-regular graph with $2n$ vertices. Suppose that the graph doesn't have a perfect matching. Then the graph is the union of two $n$-clique. Where can I find this lemma's proof? Proof. If $G$ is connected, then it has path length $2n-2$ by Exercise 4 in #5's comment. Let the vertice not in this path be $v_0$, and number the rest of verices as $v_1, ..., v_{2n-1}$ along the path. If $v_0$ is connected to $v_{2i-1}$ for some $1\leq i\leq n$, we can make perfect matching by following way; cut two edges in path incident to $v_{2i-1}$, then we can make two paths of odd length. Alternately color each of them by black and white(using more black color in each), and color the edge connecting $v_{2i-1}, v_0$ by black color. Then we get black colored perfect matching, which is contradict to our problem condition. So $v_0$ must connected to ${v_2, v_4..., v_{2n-2}}$. If we replace path $v_{2i-2}-v_{2i-1}-v_{2i}$ by $v_{2i-2}-v_0-v_{2i}$($2\leq i\leq n-1$) and use same argument as above, we can prove $v_{2i-1}$ is connected to ${v_2, v_4..., v_{2n-2}}$, too. Same argument works for $v_1, v_{2n-1}$ by replacing $v_1-v_2$, $v_{2n-2}-v_{2n-1}$ by $v_0-v_2$, $v_{2n-2}-v_0$ each. This means ${v_2, v_4..., v_{2n-2}}$ is all connected to ${v_0, v_1, v_3..., v_{2n-1}}$. Since $v_2$ has at least $n+1$ neighbor, we get contradiction. So our graph has more than $1$ connected component. Since there are $2n$ vertice and each component has size$\geq n$, it must have two connected components of size exactly $n$. It's easy to show that this two are $K_n$. Comment. This problem essentially asks for the smallest $k$ that $k-$regular graph on $2n$ vertice always have perfect matching. As we can see in proof in #3, the answer is $n$ for odd $n$, and $n-1$ for even $n$. We can prove it easily by above lemma and dirac's theorem.