Prove that either the average of the numbers $a_{1}, a_{2},..., a_{n}$ Let $a_{1}, a_{2},..., a_{2n}$ be positive real numbers such that $a_{j}a_{n+j}=1$ for the values $j=1,2,...,n$
($a$)  Prove that either the average of the numbers $a_{1}, a_{2},..., a_{n}$ is at least 1 or the average of the numbers $a_{n+1}, a_{n+2},..., a_{2n}$ is at least $1$
($b$)  Assuming that $n\geq 2$ prove that there exist two distinct numbers $j,k$ in the set $\left \{ 1,2,..,2n \right \}$ such that $$\left | a_{j}-a_{k} \right |&lt; \frac{1}{n-1}$$

I tried to write part $(a)$ like this $$\frac{\sum_{i=1}^{n}a_{i}}{n}+\frac{\sum_{i=1}^{n}a_{n+i}}{n}$$
but I don't know if it's true,
the part $(b)$m I think there will be exactly $n$ values of $i$ for which $0&lt; a_{i}\leq 1$
and by the way, this is Olympiad question, any help will thankful.
Hints:

a) It's enough to show that the average of all the numbers is at least $1$. You can do this with a famous inequality...

b) At least $n$ of the numbers must lie in $(0,1]$. Use the pigeonhole principle, with intervals of length $\frac{1}{n-1}$ as your pigeonholes.
Ok, I will complete with your solution,

Part $(a)$ $$\frac{\sum_{i=1}^{n}a_{i}}{n}+\frac{\sum_{i=1}^{n}a_{n+i}}{n}$$
$$\Rightarrow  \frac{1}{n}\sum_{i-1}^{n}\left ( a_{i}+\frac{1}{a_{i}} \right )\geq 2$$

Therefore either $$\frac{\sum_{i=1}^{n}a_{i}}{n}\geq 1$$ or $$\frac{\sum_{i=1}^{n}a_{n+i}}{n}\geq 1$$

DONE

(b) there will be exactly $n$ values of $i$ for which $0&lt; a_{i}\leq 1$. Partition $(0,1]$ into $n-1$ intervals $(\frac{a}{n-1}, \frac{a+1}{n-1}]$ 
for $a=0,1,...,n-2$. 

Therefore, by the Dirichlet principle, there are distinct $j,k$ such that $a_{j}, a_{k}$ are in the same half-open interval of length $1/(n-1)$ and $$\left | a_{j}-a_{k} \right |&lt; \frac{1}{n-1}$$

AND WE'RE DONE