Let’s color the $(1986\times 2)$ positions in alternating black and white. Per given conditions (there are $1$ number between two $1$’s, $2$ numbers between two $2$’s, and so on), there are even number of intervals between two even numbers. So, a pair of even numbers must occupy one black and one white positions. Similarly, a pair of odd numbers must be separated by an odd number of spaces, thus they must occupy two positions of the same color.

Now, there are $993$ pair of even numbers, therefore they will occupy $993$ black and $993$ white positions. Let B and W denote the number of paired odd numbers that occupy black and white positions, respectively. Then, it must hold that $B+W =993$. Because each pair contains two numbers, the number of black positions and white positions occupied by these odd numbers will be $2B$ and $2W$, respectively.

Now, the total number of black positions equal $(993+2B)$ and the total number of white positions is $(993+2W)$. Clearly, they must be equal which means $$993+2B =993+2W$$ However, because $B+W =993$ is odd, it must hold that $B\ne W$. Hence the above equality will never hold. It follows that the answer to the original question is no.