Problem 1219

1. **Define the sequence of primes and the sum $s_n$:**
Let $\{p_1, p_2, p_3, \dots\} = \{2, 3, 5, \dots\}$ be the ordered set of primes. For every integer $n \geq 3$, let $s_n$ be the sum of all primes strictly less than $n$.

2. Assume the contrary:
Suppose there are finitely many integers $n \geq 3$ such that $s_n$ is coprime to $n$. This implies there exists some integer $N$ such that for all primes $p > N$, $p \mid s_p$.

3. **Express $s_p$ in terms of $p$:**
For primes $p_{k+1} > N$, there exist positive integers $c_i$ such that $c_i p_{k+i} = s_{k+i-1}$ for all $i \geq 1$. This implies:
$$c_{i+1} p_{k+i+1} = (c_i + 1) p_{k+i}$$

4. **Analyze the sequence $c_i$:**
Since $p_{k+i} < p_{k+i+1}$, the equation $c_{i+1} p_{k+i+1} = (c_i + 1) p_{k+i}$ implies:
$$c_{i+1} < c_i + 1$$
Therefore, $c_{i+1} \leq c_i$. Since $c_i$ is a sequence of positive integers, it must eventually become constant. Let $c_i = C$ for sufficiently large $i$.

5. Derive a contradiction:
For large $t$, taking $i = t - k$ and $t - k + 1$ in the equation $c_{i+1} p_{k+i+1} = (c_i + 1) p_{k+i}$ gives:
$$C p_{t+1} = (C + 1) p_t$$
and
$$C p_{t+2} = (C + 1) p_{t+1}$$
Combining these, we get:
$$C^2 p_{t+2} = (C + 1)^2 p_t$$
This implies:
$$p_{t+1}^2 = p_{t+2} p_t$$
which is a contradiction because the sequence of primes is strictly increasing and $p_{t+1}^2 \neq p_{t+2} p_t$.

6. Conclusion:
Since our assumption leads to a contradiction, there must be infinitely many integers $n \geq 3$ such that $s_n$ is coprime to $n$.

$\blacksquare$