The claimed maximal value is achieved at
$$\begin{gathered} a_{1}=a_{2}=\cdots=a_{2016}=1, \quad a_{2017}=\frac{a_{2016}+\cdots+a_{0}}{2017}=1-\frac{1}{2017}, \\ a_{2018}=\frac{a_{2017}+\cdots+a_{1}}{2017}=1-\frac{1}{2017^{2}}. \end{gathered}$$
Now we need to show that this value is optimal. For brevity, we use the notation
$$S(n, k)=a_{n-1}+a_{n-2}+\cdots+a_{n-k} \quad \text{for nonnegative integers } k \leqslant n.$$
In particular, $S(n, 0)=0$ and $S(n, 1)=a_{n-1}$. In these terms, for every integer $n \geqslant 2$ there exists a positive integer $k \leqslant n$ such that $a_{n}=S(n, k) / k$. For every integer $n \geqslant 1$ we define
$$M_{n}=\max _{1 \leqslant k \leqslant n} \frac{S(n, k)}{k}, \quad m_{n}=\min _{1 \leqslant k \leqslant n} \frac{S(n, k)}{k}, \quad \text{and} \quad \Delta_{n}=M_{n}-m_{n} \geqslant 0.$$
By definition, $a_{n} \in [m_{n}, M_{n}]$ for all $n \geqslant 2$; on the other hand, $a_{n-1}=S(n, 1) / 1 \in [m_{n}, M_{n}]$. Therefore,
$$a_{2018}-a_{2017} \leqslant M_{2018}-m_{2018}=\Delta_{2018},$$
and we are interested in an upper bound for $\Delta_{2018}$.

Also by definition, for any $n > 2$, we have $\Delta_{n} \leqslant \frac{n-1}{n} \Delta_{n-1}$.

Proof. Choose positive integers $k, \ell \leqslant n$ such that $M_{n}=S(n, k) / k$ and $m_{n}=S(n, \ell) / \ell$. We have $S(n, k)=a_{n-1}+S(n-1, k-1)$, so
$$k\left(M_{n}-a_{n-1}\right)=S(n, k)-k a_{n-1}=S(n-1, k-1)-(k-1) a_{n-1} \leqslant (k-1)\left(M_{n-1}-a_{n-1}\right),$$
since $S(n-1, k-1) \leqslant (k-1) M_{n-1}$. Similarly, we get
$$\ell\left(a_{n-1}-m_{n}\right)=(\ell-1) a_{n-1}-S(n-1, \ell-1) \leqslant (\ell-1)\left(a_{n-1}-m_{n-1}\right).$$
Since $m_{n-1} \leqslant a_{n-1} \leqslant M_{n-1}$ and $k, \ell \leqslant n$, the obtained inequalities yield
$$\begin{array}{ll} M_{n}-a_{n-1} \leqslant \frac{k-1}{k}\left(M_{n-1}-a_{n-1}\right) \leqslant \frac{n-1}{n}\left(M_{n-1}-a_{n-1}\right) \quad \text{and} \\ a_{n-1}-m_{n} \leqslant \frac{\ell-1}{\ell}\left(a_{n-1}-m_{n-1}\right) \leqslant \frac{n-1}{n}\left(a_{n-1}-m_{n-1}\right). \end{array}$$
Therefore,
$$\Delta_{n}=\left(M_{n}-a_{n-1}\right)+\left(a_{n-1}-m_{n}\right) \leqslant \frac{n-1}{n}\left(\left(M_{n-1}-a_{n-1}\right)+\left(a_{n-1}-m_{n-1}\right)\right)=\frac{n-1}{n} \Delta_{n-1}.$$

Back to the problem, if $a_{n}=1$ for all $n \leqslant 2017$, then $a_{2018} \leqslant 1$ and hence $a_{2018}-a_{2017} \leqslant 0$. Otherwise, let $2 \leqslant q \leqslant 2017$ be the minimal index with $a_{q}<1$. We have $S(q, i)=i$ for all $i=1,2, \ldots, q-1$, while $S(q, q)=q-1$. Therefore, $a_{q}<1$ yields $a_{q}=S(q, q) / q=1-\frac{1}{q}$.

Now we have $S(q+1, i)=i-\frac{1}{q}$ for $i=1,2, \ldots, q$, and $S(q+1, q+1)=q-\frac{1}{q}$. This gives us
$$m_{q+1}=\frac{S(q+1,1)}{1}=\frac{S(q+1, q+1)}{q+1}=\frac{q-1}{q} \quad \text{and} \quad M_{q+1}=\frac{S(q+1, q)}{q}=\frac{q^{2}-1}{q^{2}}$$
so $\Delta_{q+1}=M_{q+1}-m_{q+1}=(q-1) / q^{2}$. Denoting $N=2017 \geqslant q$ and using Claim 1 for $n=q+2, q+3, \ldots, N+1$ we finally obtain
$$\Delta_{N+1} \leqslant \frac{q-1}{q^{2}} \cdot \frac{q+1}{q+2} \cdot \frac{q+2}{q+3} \cdots \frac{N}{N+1}=\frac{1}{N+1}\left(1-\frac{1}{q^{2}}\right) \leqslant \frac{1}{N+1}\left(1-\frac{1}{N^{2}}\right)=\frac{N-1}{N^{2}}$$
as required.

Comment 1. One may check that the maximal value of $a_{2018}-a_{2017}$ is attained at the unique sequence, which is presented in the solution above.

Comment 2. An easier question would be to determine the maximal value of $\left|a_{2018}-a_{2017}\right|$. In this version, the answer $\frac{1}{2018}$ is achieved at
$$a_{1}=a_{2}=\cdots=a_{2017}=1, \quad a_{2018}=\frac{a_{2017}+\cdots+a_{0}}{2018}=1-\frac{1}{2018}.$$
To prove that this value is optimal, it suffices to notice that $\Delta_{2}=\frac{1}{2}$ and to apply Claim 1 obtaining
$$\left|a_{2018}-a_{2017}\right| \leqslant \Delta_{2018} \leqslant \frac{1}{2} \cdot \frac{2}{3} \cdots \frac{2017}{2018}=\frac{1}{2018}.$$