a.
Let $x_1, x_2, \dots, x_n$ be non-negative integers satisfying the conditions from the statement. We put $M = \max_{1 \le i \le n} x_i$. If $x_j = M$ then
$$x_1x_2 + x_2x_3 + \dots + x_{n-1}x_n \le x_1x_j + x_2x_j + \dots + x_{j-1}x_j + x_jx_{j+1} + x_jx_{j+2} + \dots + x_jx_n = x_j(2011-x_j) = M(2011-M) \le 1005 \cdot 1006.$$
Indeed, the last inequality comes to $(M-1005)(M-1006) \ge 0$ which is true for any integer $M$. The largest possible value is $1005 \cdot 1006$ because this value can be obtained by choosing, for example, $x_1 = 1005$, $x_2 = 1006$ and $x_k = 0$ for $k \ge 3$.

b.
For $n=2$ we have $x_1x_2 = 1005 \cdot 1006 \Leftrightarrow x_1(2011-x_1) = 1005 \cdot 1006 \Leftrightarrow (x_1-1005)(x_1+1006) = 0 \Leftrightarrow (x_1,x_2) \in \{(1005, 1006), (1006, 1005)\}$.

For $n=3$, $x_1x_2 + x_2x_3 = 1005 \cdot 1006 \Leftrightarrow x_2(x_1+x_3) = 1005 \cdot 1006$ comes, as above, to $x_2 = 1005$, $x_1+x_3 = 1006$ or $x_2 = 1006$, $x_1+x_3 = 1005$. We obtain $(x_1, x_2, x_3) \in \{(k, 1005, 1006-k) \mid k = 0, 1, \dots, 1006\} \cup \{(k, 1006, 1005-k) \mid k = 0, 1, \dots, 1005\}$.

For $n \ge 4$, we denote by $j$ the smallest index for which $x_j > 0$. Then, replacing $x_j$ by $0$ and $x_{j+2}$ by $x_{j+2} + x_j$, increases the value of the sum by $x_jx_{j+3}$. Using this remark it is easy to see that if $x_1x_2 + x_2x_3 + \dots + x_{n-1}x_n = 1005 \cdot 1006$, then at most three of the terms can be non-zero. We obtain $(x_1, \dots, x_n) \in \{(0, \dots, 0, k, 1005, 1006-k, 0, \dots, 0) \mid k = 0, 1, \dots, 1006\} \cup \{(0, \dots, 0, k, 1006, 1005-k, 0, \dots, 0) \mid k = 0, 1, \dots, 1005\}$, where the group of the three non-zero components can be located anywhere.