The greatest such $C$ is $1010$. First, to see that this cannot be improved, take $(a_1, a_2, \cdots, a_{2019}) = (x, 1, x+y, 1+y, x+2y, 1+2y, x+3y, \cdots, x+1008y, 1+1008y, 1+\epsilon),$ where $x = \epsilon^2, y = \epsilon^3$ and $\epsilon$ is tiny. Now, we will show that the quantity in question is indeed greater than $1010$. For convenience, let's allow $a_i$ to be $0$ (now we need to show that the quantity is greater than or equal to $1010$). Firstly, observe that if we decrease all of the $a_i$'s by a certain amount, say $x$, then each of the fractions in the sum decreases. Hence, we can clearly assume that $a_1 = 0$ (let $a_1$ be the smallest one). Now, for $i \neq 1$, let $f(i)$ be the number among $i+1, i+2$ for which $a_{f(i)}$ is bigger (can't be same, since $a_i$'s are distinct), where indices are cycled modulo $2019$. Consider the sequence $2, f(2), f(f(2)), \cdots$. Observe that $f(i) - i$ is $1$ or $2$ for all $i$, by definition. Therefore, we must eventually have some $k$ such that $f^{(k)}(2) = 2018$ or $2019$ (it can't "jump over" both these values). Since $f(2019) = 2$ and $f(2018) = 2019$ (since $a_1 = 0$), we have that this sequence is periodic in either case. Again, since it increases by at most $2$ each time and increased by a total of $2019$, we have that the period is at least $1010$. Call the period $p \geq 1010$. Now, the key observation is that $\frac{a_i}{|a_{i+1} - a_{i+2}|} \geq \frac{a_i}{a_{f(i)}}$ for all $i$. Applying this for $i = 2, f(2), f(f(2)), \cdots, f^{(p-1)}(2)$ and summing, we know that: $$\sum_{i= 1}^{2019} \frac{a_i}{|a_{i+1} - a_{i+2}|} \geq \sum_{i=0}^{p-1} \frac{a_{f^{(i)}(2)}}{a_{f^{(i+1)}(2)}} \geq p,$$ where the last $\geq$ follows by $AM-GM$. In conclusion, the sum in question is at least $1010$. Now it's pretty easy to check that equality cannot possibly hold, so we're done.

The greatest such $C$ is $1010$. First, to see that this cannot be improved, take $(a_1, a_2, \cdots, a_{2019}) = (x, 1, x, 1, x, 1, x \cdots, x, 1, 1+\epsilon),$ where $x = \epsilon^2$ and $\epsilon$ is tiny. Now, we will show that the quantity in question is indeed greater than $1010$. For convenience, let's allow $a_i$ to be $0$ (now we need to show that the quantity is greater than or equal to $1010$). Firstly, observe that if we decrease all of the $a_i$'s by a certain amount, say $x$, then each of the fractions in the sum decreases. Hence, we can clearly assume that $a_1 = 0$ (let $a_1$ be the smallest one). Now, for $i \neq 1$, let $f(i)$ be the number among $i+1, i+2$ for which $a_{f(i)}$ is bigger (if $a_{i+1} = a_{i+2}$ the problem is broken, so we ignore this case), where indices are cycled modulo $2019$. Consider the sequence $2, f(2), f(f(2)), \cdots$. Observe that $f(i) - i$ is $1$ or $2$ for all $i$, by definition. Therefore, we must eventually have some $k$ such that $f^{(k)}(2) = 2018$ or $2019$ (it can't "jump over" both these values). Since $f(2019) = 2$ and $f(2018) = 2019$ (since $a_1 = 0$), we have that this sequence is periodic in either case. Again, since it increases by at most $2$ each time and increased by a total of $2019$, we have that the period is at least $1010$. Call the period $p \geq 1010$. Now, the key observation is that $\frac{a_i}{|a_{i+1} - a_{i+2}|} \geq \frac{a_i}{a_{f(i)}}$ for all $i$. Applying this for $i = 2, f(2), f(f(2)), \cdots, f^{(p-1)}(2)$ and summing, we know that: $$\sum_{i= 1}^{2019} \frac{a_i}{|a_{i+1} - a_{i+2}|} \geq \sum_{i=0}^{p-1} \frac{a_{f^{(i)}(2)}}{a_{f^{(i+1)}(2)}} \geq p,$$ where the last $\geq$ follows by $AM-GM$. In conclusion, the sum in question is at least $1010$. Now it's pretty easy to check that equality cannot possibly hold, so we're done.

(this step needs some courage)