# Dividing first $n$ primes into two sets with equal sum

Let $N$ be a positive integer. Does there always exist $n>N$ such that the first $n$ primes can be divided into two sets with equal sum?

If $n$ is even, the sum of the first $n$ primes is odd, so we cannot perform the division. But if $n$ is odd, it might be possible. For instance, $2+3=5, 2+5+7=3+11, 2+3+11+13=5+7+17$.

He proves the result that, given the first $n$ primes $\{p_1,\ldots,p_n\}$, we can find some $e_1,\ldots,e_n$ with $e_i = \pm 1$ such that $$\left\vert \sum_{i=1}^n e_i p_i \right\vert \leq 1$$ A simple parity argument tells us that if $n$ is odd, this sum must be $0$. We then partition the primes into those with coefficient $+1$ and $-1$, and lo, a partition of the first $n$ primes with equal sum.
It follows that for any $N$, we can choose any odd $n>N$ and find such a partition of the primes. (Which seems stronger than merely some $n>N$ exists.)