We chose $n + 2$ numbers from the set $\{1,2,....3n\}$ . Prove that there are always two among the chosen numbers whose difference is more than $n$ but less than $2n$.
Though I can understand it by taking examples but I really struggle when it comes to prove. Also is there really good book that will structure my thinking so that I can proof these type of questions.
Like Alexey suggested you can consider the remainders of division by $n+1$ of these $n+2$ numbers.
Since the different possibile remainders are only $n+1$ and you have $n+2$ different possible solutions you know, thanks to the pidegeonhole principle, that at least $2$ different numbers, $x$ and $y$, have the same remainder.
You can suppose that $y>x$ and you know that $n+1$ divides $y-x$. Since they are different you know that $y-x= n+1$ or $y-x=2n+2$. In the first case you have found the numbers you need.
In the second case, you know that $x \le n$ and $y= 2n+2+x \ge 2n$.
Now we can consider the other $n$ numbers. The only numbers which difference from $x$ and from $y=2n+x+2$ are not between $n$ and $2n$ are the ones below $x+3$ and above $x+2n-3$ and $3n$. These numbers, excluding $x$ and $y$ are $x+3 + (3n-(2n+x-3)) -2 = n+4$ but you can't take them all. You can freely take all the numbers below $x$ and above $y$, and you get $x + (3n - (2n+x+2)) = n-2 $ numbers but you can't choose all of the $6$ different numbers remaining because, for example, if you choose $x+3$ you can't choose $ 2n+x-3$ or $2n+x-2$ or $2n+x-1$ .
You have to choose $4$ numbers between these $6$, this are only $15$ cases and you notice that you can't do that.