probably overkill but: Choose $2019 < a_1 = p_1 < a_2 = p_2 < \dots < a_{2017} = p_{2017}$ to be arbitrary fixed primes. Suppose we want to write $m = a_1 + \dots + a_{2019}$ with given conditions. Let $n = m - (p_1+\dots+p_{2017})$. For each $1 \leq i \leq 2017$, let $x_i$ be such that $x_i \not \equiv 0 \pmod{p_i}$ and $x_i \not \equiv n \pmod{p_i}$. By CRT, there is $\ell < p_1p_2\dots p_{2017}$ such that $\ell \equiv x_i \pmod{p_i}$ for all $1 \leq i \leq 2017$, let $t = p_1p_2\dots p_{2017}$. By prime number theorem on arithmetic progressions, for large $n$ there are at least two primes in $\{kt+\ell : k=\lceil \frac{n}{2t} \rceil ,\dots,\lfloor \frac{2n}{3t} \rfloor \}$, these primes can't both divide $n$ for large $n$, hence one of them is coprime with $n$, call it $p$. Choosing $a_{2018} = n-p , a_{2019} = p$ works.