For every $n > 0$, let $b_n = \ln a_n$. We can reduce the problem to: Given 2017 reals $b_1, b_2, \dots, b_{2017}$. For every $n > 2017$, let
$$b_n = \max\{b_{i_1} + b_{i_2} + b_{i_3} \mid i_1 + i_2 + i_3 = n, 1 \le i_1 \le i_2 \le i_3 \le n-1\}.$$
Prove there exist positive integers $m \le 2017$ and $N > 4m$ such that $b_n + b_{n-4m} = 2b_{n-2m}$ for every $n > N$.

Consider $\ell$ ($1 \le \ell \le 2017$) such that $\frac{b_\ell}{\ell} = \max\left\{\frac{b_i}{i} \mid 1 \le i \le 2017\right\}$. We have the following statement:

Claim 1. For every $n \in \mathbb{Z}^+$:
$$\frac{b_n}{n} \le \frac{b_\ell}{\ell}.$$
Proof. We prove by induction on $n$. The statement is clearly true for $n \le 2017$ by the definition of $\ell$. We consider the case $n > 2017$. Assume it is true for all $k < n$. From the definition of $b_n$, there exist $j_1, j_2, j_3 \in \mathbb{N}^*$ satisfying $j_1 + j_2 + j_3 = n$ such that
$$b_n = b_{j_1} + b_{j_2} + b_{j_3}.$$
Using the induction hypothesis, we have
$$b_n \le j_1 \cdot \frac{b_\ell}{\ell} + j_2 \cdot \frac{b_\ell}{\ell} + j_3 \cdot \frac{b_\ell}{\ell} = n \cdot \frac{b_\ell}{\ell},$$
so $\frac{b_n}{n} \le \frac{b_\ell}{\ell}$. Therefore the statement is also true for $n$. The claim has proven.

Now, for every positive integer $n$, let $c_n = n b_\ell - \ell b_n$ then from the above claim we have $c_n \ge 0$ for every $n$. We also have, for $n \ge 2017$:
$$\begin{aligned} c_{n+2\ell} &= (n+2\ell)b_\ell - \ell b_{n+2\ell} \\ &\le (n+2\ell)b_\ell - \ell(b_n + b_\ell + b_\ell) \\ &= n b_\ell - \ell b_n = c_n. \end{aligned}$$
Hence
$$c_{n+2k\ell} \le c_{n+2(k-1)\ell} \le \dots \le c_n, \quad \forall n \ge 2017, k \ge 1.$$
Let $x$ be the smallest positive integer such that $2x\ell > 2017$ and set
$$M = \max\{c_i \mid 1 \le i \le 4x\ell - 1\}.$$
Then, for every $n > 2x\ell$, set $n = 2kx\ell + r$ (with $0 \le r < 2x\ell$), we have
$$c_n \le c_{r+2(k-1)x\ell} \le c_{r+2(k-2)x\ell} \le \dots \le c_{r+2x\ell} \le M.$$
We have the following statements:

Claim 2. For every positive integer $n$, there exist natural numbers $s_1, s_2, \dots, s_{2017}$ such that
$$c_n = s_1 c_1 + s_2 c_2 + \dots + s_{2017} c_{2017}.$$
Proof. We prove by induction on $n$. The statement is clearly true for $n \le 2017$ (just choose $s_n = 1$ and $s_i = 0$ with $i \ne n$). We just need to consider the case $n > 2017$ is sufficient. Suppose it is true for $k < n$. From the definition of $b_n$, we deduce
$$c_n = \min\{c_{i_1} + c_{i_2} + c_{i_3} \mid 1 \le i_1 \le i_2 \le i_3 \le n-1, i_1 + i_2 + i_3 = n\}, \quad \forall n > 2017.$$
Thus there exist $j_1, j_2, j_3 \in \mathbb{N}^*$ satisfying $j_1 + j_2 + j_3 = n$ such that
$$c_n = c_{j_1} + c_{j_2} + c_{j_3}.$$
By the induction hypothesis, there exist natural numbers
$$u_1, u_2, \dots, u_{2017}, v_1, v_2, \dots, v_{2017}, w_1, w_2, \dots, w_{2017}$$
such that
$$\begin{align*} c_{j_1} &= u_1 c_1 + u_2 c_2 + \dots + u_{2017} c_{2017}, \\ c_{j_2} &= v_1 c_1 + v_2 c_2 + \dots + v_{2017} c_{2017}, \\ c_{j_3} &= w_1 c_1 + w_2 c_2 + \dots + w_{2017} c_{2017}. \end{align*}$$
Therefore, we have
$$c_n = s_1 c_1 + s_2 c_2 + \dots + s_{2017} c_{2017},$$
where $s_i = u_i + v_i + w_i$ for $1 \le i \le 2017$. Hence the statement is also true for $n$. The claim has proven.

Claim 3. The sequence $(c_n)$ get only finite values.

Proof. This statement is inferred directly from the boundedness of $c_n$ and the result of above claim.

Now, since $c_n$ get only finite values and
$$c_{n+2kx\ell} \le c_{n+2(k-1)x\ell} \le \dots \le c_n$$
for every $n \ge 2017$ so there exists $N_1$ big enough such that $c_n = c_{n-2x\ell}$ for every $n > N_1$. Then, we have
$$n b_{\ell} - \ell b_n = (n - 2x\ell) b_{\ell} - \ell b_{n-2x\ell}, \quad \forall n > N_1,$$
or
$$b_n = 2b_{\ell} + b_{n-2x\ell}, \quad \forall n > N_1.$$
Then, we conclude that
$$b_{n-2x\ell} = 2b_{\ell} + b_{n-4x\ell}, \quad \forall n > N_1 + 2x\ell.$$
From two above results, we get
$$b_n + b_{n-4x\ell} = 2b_{n-2x\ell}, \quad \forall n > N_1 + 2x\ell.$$
Choose $N = N_1 + 2x\ell$ and $m = x\ell$, we have the desired result.