Cute problem. As expected, the answer is yes. Let $\{p_1,p_2,p_3,\dots\}=\{2,3,5,\dots\}$ be the ordered set of primes. Supposing that there were finitely many $n$ with the given property, there is an $N$ such that $p\mid s_p$ for all primes $p>N$. Then taking $p_{k+1}>N$, we know that there exist positive integers $c_i$ such that $c_{i}p_{k+i}=s_{k+i-1}$ for all $i\geq 1$, which implies that \[ c_{i+1}p_{k+i+1} = (c_i+1)p_{k+i}.\] Noting that $p_{k+i}<p_{k+i+1}$, the above implies that $c_{i+1} < c_i+1$, so $c_{i+1}\leq c_i$. Since $c_i$ is a sequence of positive integers, it then must be eventually constant. If $c_i=C$ for sufficiently large $i$, taking $i=t-k$ and $t-k+1$ for large $t$ in the above equation gives $Cp_{t+1} =(C+1)p_{t}$ and $Cp_{t+2} = (C+1)p_{t+1}$, which together imply $p_{t+1}^2=p_{t+2}p_t$, a clear contradiction. Thus there are infinitely many primes $p$ such that $\gcd(p,s_p)=1$, and we're done.

Remark

Here is my solution (a bit shorter) Assume there are finitely many such $n$. Let $p_i$ be the $i$th prime number. Then there exists some $N$ such that for any $n\geq N$, $\gcd(p_n,s_{p_n})\neq 1$. Therefore, $p_n\mid s_{p_n}$. Let $s_{p_n}=k\cdot p_n$. Note that we also have $p_{n+1}\mid s_{p_{n+1}}=s_{p_n}+p_n=(k+1)p_n.$ Observe that because $p_{n+1}$ and $p_n$ are different prime numbers then we have $p_{n+1}\mid (k+1).$ Therefore \[p_{n+1}\leq k+1\iff p_{n+1}p_n\leq p_n(k+1)=p_1+p_2+...+p_{n+1}\leq(n+1)p_{n+1}\]which gives us $p_n\leq n+1$ which is a contradiction.