Solution:

Write each coefficient as $a_i = 10b_i + c_i$ with digits $b_i, c_i \in \{0, 1, \ldots, 9\};$ every integer $0 \le a_i \le 99$ splits this way uniquely. Setting $m = b_3 b_2 b_1 b_0$ and $n = c_3 c_2 c_1 c_0$ (read as base-$10$ numbers), the condition becomes $$2010 = 10m + n.$$

Conversely, any nonnegative integers $m, n$ with $10m + n = 2010$ satisfy $m \le 201$ and $n \le 2010,$ so each has at most four digits; those digits recover the $b_i$ and $c_i,$ hence the $a_i.$ So representations correspond exactly to choices of $m \in \{0, 1, \ldots, 201\}$ with $n = 2010 - 10m,$ and $N = 202.$