Solution:
It is possible, as long as the sum $S$ desired is a multiple of $48$, with $N = S / 6$, which in this case is $3360336$, and the $a_{k}$ repeats the 8-term pattern $-1, 1, 1, -1, 1, -1, -1, 1$.

Use the observation that if $f(x)$ is a degree-$k$ polynomial, then for any constant $h$, the difference $f(x+h)-f(x)$ will be a degree-$(k-1)$ polynomial. If we iterate this process three times, we can find a way to manipulate consecutive cubes to always get a constant.

More precisely, let
$$c_{0}, c_{1}, c_{2}, c_{3}, c_{4}, c_{5}, c_{6}, c_{7}, \ldots$$
be consecutive cubes. In other words, $c_{m} = (m+u)^{3}$ for some fixed starting integer $u$. Then the differences
$$c_{1}-c_{0}, \quad c_{3}-c_{2}, \quad c_{5}-c_{4}, \quad c_{7}-c_{6}, \ldots$$
will be quadratic functions; i.e., if we define $a_{m} := c_{m+1} - c_{m}$, then $a_{m}$ is a quadratic function of $m$ (depending on the parameter $u$, as well), and the differences are $a_{0}, a_{2}, a_{4}, a_{6}, \ldots$ Continuing, we see that the differences
$$a_{2}-a_{0}, \quad a_{6}-a_{4}, \ldots$$
will be a linear sequence; i.e., $b_{m} := a_{m+2} - a_{m}$ is a linear function of $m$ (with parameter $u$), and our differences are $b_{0}, b_{4}, \ldots$ Finally, the sequence
$$b_{4}-b_{0}, \quad b_{12}-b_{8}, \ldots$$
is constant, no matter what the parameter $u$ equals! We have
$$b_{4}-b_{0} = a_{6}-a_{4} - (a_{2}-a_{0}) = a_{6}-a_{4}-a_{2}+a_{0} = c_{7}-c_{6}-c_{5}+c_{4}-c_{3}+c_{2}+c_{1}-c_{0},$$
and since this is constant, we can compute it using any value of $u$. Taking $u = -3$, the constant must equal
$$4^{3}-3^{3}-2^{2}+1^{3}-0^{3}+(-1)^{3}+(-2)^{3}-(-3)^{3}=48.$$
In other words, if we define
$$s_{u} := -u^{3} + (u+1)^{3} + (u+2)^{3} - (u+3)^{3} + (u+4)^{3} - (u+5)^{3} - (u+6)^{3} + (u+7)^{3},$$
then $s_{u} = 48$ for all values of $u$.

Since $20162016 = 420042 \cdot 48$, we can easily write $20162016$ as a sum of $420042$ 8-element sum/differences of $420042 \cdot 8 = 3360336$ consecutive cubes:
$$\begin{aligned} 20162016 = \sum_{k=0}^{420041} s_{8k+1} = &\ -1^{3} + 2^{3} + 3^{3} - 4^{3} + 5^{3} - 6^{3} - 7^{3} + 8^{3} \\ & -9^{3} + 10^{3} + 11^{3} - 12^{3} + 13^{3} - 14^{3} - 14^{3} + 15^{3} \\ & \cdots \\ & -3360329^{3} + 3360330^{3} + \cdots -3360335^{3} + 3360336^{3}. \end{aligned}$$