Answer: a) impossible; b) possible.

a) Consider square $n \times n$, for positive integer $n$. Let us denote the most left column by $1$, adjacent column by $2$, and so on, the most right column by $n$. For any cell $c$ we denote function $w(c) = b$, provided that cell $c$ is located in the column $b$. For every token $T$ in the same way let $w(T) = b$ provided that token $T$ is placed on the cell in the column $b$ at the moment. Finally, by $W$ we denote the sum of $w(T)$ for all tokens $T$ on the board. It is clear that after every round $W$ is not changed. At the end of our process we must have
$$W = n + 2n + \dots + n^2 = n \cdot (1 + 2 + \dots + n) = \frac{n^2(n+1)}{2}.$$
At the beginning of our process $W = l \cdot n^2$ provided that all tokens are placed at the cell in $l$. And for $n = 2016$ we get a contradiction.

b) Let us consider the case $n = 2017$. It follows from previous arguments that the statement of the problem could be true only in the case when all tokens are placed on the central cell at the beginning. In this case we provide the algorithm. Let us note that in every cell we write the number of tokens placed there. Using induction, we prove the following statements

Statement 1. For every positive integer $n$ in a row of length $2n+1$ from the position
$0; 0; 0; \dots; 0; 2n+1; 0; \dots; 0; 0$
we can get the position
$1; 1; 1; \dots; 1; 1; 1; \dots; 1; 1$

Statement 2. For every positive integer $n$ in a row of length $2n+1$ from the position
$0; 1; 1; \dots; 1; 3; 1; \dots; 1; 1$
we can get the position
$1; 1; 1; \dots; 1; 1; 1; \dots; 1; 1$

Proof. It is clear for $n=1$. Suppose that for $n=k-1$ statements are true, let us prove them for $n=k$. At the beginning, we have $2k+1$ tokens on the central cell. Two of them we will not move. The rest $2k-1$ tokens we can move (according to the induction hypothesis) to the position
$0; 1; 1; \dots; 1; 1; 3; 1; 1; \dots; 1; 1$
Now consider the following replacing
$0; 1; 1; \dots; 1; 1; 2; 1; 2; 1; 1; \dots; 1; 1$
$0; 1; 1; \dots; 1; 2; 0; 3; 0; 2; 1; \dots; 1; 1$
$0; 1; 1; \dots; 2; 0; 1; 3; 1; 0; 2; \dots; 1; 1$
$\dots$
$0; 1; 2; 0; 1; \dots; 1; 1; 1; 3; 1; 1; 1; \dots; 1; 0; 2; 1$
$0; 2; 0; 1; 1; \dots; 1; 1; 1; 3; 1; 1; 1; \dots; 1; 1; 0; 2$
$1; 0; 1; 1; 1; \dots; 1; 1; 1; 3; 1; 1; 1; \dots; 1; 1; 1; 0$
We can apply the induction hypothesis now, and get
$1; 1; 1; \dots; 1; 1; 1; \dots; 1; 1$
Both statements are proved now.

We can easily prove the statement of the problem now. Divide all tokens into equal groups with $2017$ tokens in every group. Using our statements, we can replace tokens such that in every cell of the central column exactly one group will be placed. And then just use the statements again to replace tokens such that in every cell of every row exactly one token will be placed.