Put $1,2,....,2018$ (2018 numbers) in a row randomly and call this number $A$. Find the remainder of $A$ divided by $3$.
1. We are given the sequence of numbers \(1, 2, \ldots, 2018\) and we need to find the remainder when the number \(A\) formed by arranging these numbers in a row is divided by 3.
2. To solve this, we first consider the sum of the numbers from 1 to 2018. The sum \(S\) of the first \(n\) natural numbers is given by the formula:
   \[
   S = \frac{n(n+1)}{2}
   \]
   Substituting \(n = 2018\):
   \[
   S = \frac{2018 \cdot 2019}{2}
   \]
3. We need to find the remainder of \(S\) when divided by 3. First, we simplify the expression modulo 3:
   \[
   2018 \equiv 2 \pmod{3}
   \]
   \[
   2019 \equiv 0 \pmod{3}
   \]
   Therefore,
   \[
   S = \frac{2018 \cdot 2019}{2} \equiv \frac{2 \cdot 0}{2} \equiv 0 \pmod{3}
   \]
4. Since the sum \(S\) of the numbers from 1 to 2018 is congruent to 0 modulo 3, the sum of the digits of \(A\) will also be congruent to 0 modulo 3. This is because rearranging the numbers does not change their sum.
5. Therefore, the number \(A\) itself, when divided by 3, will leave a remainder of 0.

The final answer is \(\boxed{0}\).