2016/065) For which positive integer n does 2n devide the sum of $1^{st}$ n numbers, does 2n+1 devide the sum of $1^{st}$ n numbers

we have sum of $1^{st}$ n numbers =$\frac{n(n+1)}{2}$
for this to be divisible by 2n we should have $4n$ should devide$n(n+1)$ or 4 should devide n+1 or n should be of the form $4k-1$

as (2n+1) is co-prime to n and n+1 so for no n 2n+1 devides the sum