This Statement is Frue. Wait What?

Take any statement $P_x$ from $P_{n+1}$ to $P_{2n}$ and assume it is true. Then since $P_x$ is true, this implies that at least $x$ statements are false ( $n+1\leqslant x \leqslant 2n$ ).

If any statement from $P_{n+1}$ to $P_{2n}$ is true, statements $P_1$ to $P_n$ must all be true (can you figure out why?).

We now have at least $n+1$ true statements, in other words, we have at most $n-1$ false statements. This contradicts $P_x$ as $n-1<x$ .

Hence all the statements from $P_{n+1}$ to $P_{2n}$ are false.

Now take any statement $P_y$ from $P_1$ to $P_n$ and assume it is false. Then we have at least $n+1$ false statements, which implies that $P_{n+1}$ is true, however this is a contradiction as $P_{n+1}$ is already false.

Hence all the statements from $P_1$ to $P_n$ are true.

Therefore, the number of true statements is $n$ .

There can be 3 cases, $n$ can be true, $n+k$ can be true or $n-k$ can be true where $k$ is a positive integer.

1. $n$ are true: Since n state,ents are true, all the statements from $1$ to $n$ are true and rest are false, which implies our assumption.

2. $n+k$ are true: $n$ statements are true definitely, so $1$ to $n$ are true and $k$ statements of the rest are true. But the rest cannot be true because they can only be true when atleast $n+1$ are false or atmost $n-1$ are true. A contradiction.

3. $n-k$ are true: First $n-k$ statements are true definitely. This means $n+k$ are false, so atleast $n+1$ should be false, making $(n+1)^{th}$ statement true, which means atleast $n-k+1$ are true, which contradicts our assumption.

So the only possible case is $\boxed {n}$ statements are true.