Going around the circle, take every other point. These $N$ points must be paired up among themselves, so $N$ is even.

Label the points with the elements $0,1,\ldots,2N-1$ of $\mathbb{Z}_{2N}$. The requirement translates into $\mathbb{Z}_{2N}$ being partitioned into $N$ doubletons $\{i,j\}$ with $i+j$ (and/or, equivalently $i-j$) even. Then the sum of the elements of all these doubletons is $\sum_{N} (i+j) = \sum_{k=0}^{2N-1} k = N(2N-1)$, and being a sum of $N$ even integers, is even. Therefore $N$ must be even.