SOLVED

Is it true that, for every $n$ and $d$, there exists $k$ such that
\[d \mid p_{n+1}+\cdots+p_{n+k},\]
where $p_r$ denotes the $r$th prime?

Cedric Pilatte has observed that a positive solution to this follows from a result of Shiu [Sh00]: for any $k\geq 1$ and $(a,q)=1$ there exist infinitely many $k$-tuples of consecutive primes $p_m,\ldots,p_{m+{k-1}}$ all of which are congruent to $a$ modulo $q$.

Indeed, we apply this with $k=q=d$ and $a=1$ and let $p_m,\ldots,p_{m+{d-1}}$ be consecutive primes all congruent to $1$ modulo $d$, with $m>n+1$. If $p_{n+1}+\cdots+p_{m-1}\equiv r\pmod{d}$ with $1\leq r\leq d$ then \[d \mid p_{n+1}+\cdots +p_m+\cdots+p_{m+r-1}.\]