A well known result states that every integer $n \geq 7$ is the sum of distinct primes (we allow sums consisting just of one summand). Another way of writing this is that each $n \geq 7$ can be written as $n=\sum_{k \geq 1} a_kp_k$ where each $a_k \in \{0,1\}$ and $p_k$ is the $k$th prime.
I'm wondering if the following result is known:
For every $M \in \mathbb{N}$ there exists an $N_m \in \mathbb{N}$ such that every integer $n>N_m$ is the sum of distinct primes larger than $M$.
This theorem is suggested in this paper, but not explicitly stated. For instance, they prove the special case that every integer $>57$ is the sum of distinct primes $ \geq 13$. However, the paper proves an important lemma which may be helpful in proving the result.
I would be surprised if this problem admits a quick and easy answer, so this is more of a reference request than anything else. Although if you have a solution, you're free to share it!
$\textbf{Edit}$: The paper is behind a paywall (I initially overlooked this since I have university access by default) . The lemma is the following, quoted verbatim. Note sums of the form $\sum_{i=j}^{k} a_i$ for $j>k$ are interpreted as $0$.
$\textbf{Lemma}$ Let $\{S_i\}_{i=1}^{\infty}$ be a sequence of positive integers, and assume there exist fixed integers $r \geq 0$ and $K_0 \geq 0$ such that $(i)$ $S_{n+1} \leq S_{r+1} + \sum_{i=r+1}^{n} S_i$ for all $n \geq r$ and $(ii)$ each integer $N$ in the range $K_0 \leq N < K_0 + S_{r+1}$ can be expressed as $\sum_{i=1}^{r} a_iS_i$ for $a_i \in \{0,1\}$. Then any integer $N$ satisfying $K_0 \leq N < K_0 + S_{r+1} + \sum_{i=r+1}^{n} S_i$ for some $n \geq r$ is expressible in the form $N=\sum_{i=1}^{n} a_iS_i$ for $a_i \in \{0,1\}$.
$\textbf{Sublemma}$ The following "sub-lemma" is used to verify the first hypothesis $(i)$ in the above lemma. Namely, if a sequence $\{S_i\}_{i=1}^{\infty}$ of positive integers satisfies $S_{n+1} \leq 2S_n$ for all $n \geq r+1$, then the hypothesis $(i)$ of the above lemma holds for all $n \geq r$.
$\textbf{An illustration of the lemma's power:}$ Every $n>57$ is the sum of distinct primes each of which is at least $13$.
$\textbf{Proof:}$ Let $S_i$ be defined by $S_1 = 13, S_2 = 17, S_3 = 19, ...$ i.e., the sequence of primes $ \geq 13$. We claim that this sequence satisfies the hypotheses for lemma $1$ with $K_0=58, r=10$. To verify the first hypothesis $(i)$, this is an easy application of the sublemma and Bertrand's postulate. We now verify the second hypothesis. Note that $S_{r+1} + K_0 = 111$. We must verify that each integer $n \in [58, 110]$ is the sum distinct elements in the set $\{S_1, S_2, \cdots, S_{10}\}$; in other words, each $58 \leq n \leq 110$ is $n=\sum_{i=1}^{10} a_iS_i$ for $a_i \in \{0,1\}$. This is trivial, and can be done by hand or computer. Hence, the conclusion to lemma $1$ applies to our sequence $S_i$. Since $S_i \to \infty$, it follows that given any $N \geq 58$, there exists a large $\ell$ such that $N < K_0 + S_{r+1} + \sum_{i=r+1}^{\ell} S_i$, and hence we have that $N = \sum_{i=1}^{\ell} a_iS_i$ for $a_i \in \{0,1\}$, as desired.
The reason this approach may not be "scalable" to the general problem lies in the fact that we require computational verification for a certain range.