Three, (50 points) Let the number of all positive integers satisfying the following conditions be $N$:
(1) less than or equal to 2,012;
(2) the number of 1s in their binary representation is at least 2 more than the number of 0s.
Find the sum of the digits of $N$.


Three, from $2012=(11111011100)_{2}$, we know that the numbers satisfying the conditions have at most 11 digits in binary representation.

The first digit must be 1, so the number of $d+1$-digit numbers with exactly $k+1$ digits being 1 is $\mathrm{C}_{d}^{k}$, and condition (2) is equivalent to
$$
\begin{array}{l}
k+1 \geqslant d-k+2 \\
\Leftrightarrow k \geqslant \frac{d+1}{2} \Leftrightarrow d-k \leqslant \frac{d-1}{2} .
\end{array}
$$

First, consider the numbers in $[1,2048)$ that satisfy condition (2)
There are
$$
\begin{array}{l}
\sum_{d=1}^{10} \sum_{k=\left\{\frac{d+1}{2}\right]}^{d} \mathrm{C}_{d}^{k}=\sum_{d=1}^{10}\left[\frac{d-1}{2}\right] \\
=\sum_{a=0}^{5}\left(\sum_{i=0}^{u-1} \mathrm{C}_{2 a-1}^{i}+\sum_{i=0}^{a-1} \mathrm{C}_{2 a}^{i}\right) \\
=\sum_{a=1}^{5}\left[2^{2 a-2}+\frac{1}{2}\left(2^{2 a}-\mathrm{C}_{2 a}^{a}\right)\right] \\
=\frac{3}{4} \sum_{a=1}^{5} 4^{n}-\frac{1}{2} \sum_{a=1}^{a} \mathrm{C}_{2 a}^{a} \\
=\left(4^{5}-1\right)-\frac{1}{2} \cdot\left(\mathrm{C}_{2}^{1}+\mathrm{C}_{4}^{2}+\mathrm{C}_{6}^{3}+\mathrm{C}_{8}^{4}+\mathrm{C}_{10}^{5}\right) \\
=1023-\frac{1}{2}(2+6+20+70+252) \\
=848 \text { (numbers), }
\end{array}
$$

where $\lceil x\rceil$ represents the smallest integer not less than the real number $x$, and $[x]$ represents the largest integer not greater than the real number $x$.
Since there are 35 integers in $[2013,2048)$, and only
$$
(11111100000)_{2}=2016
$$

does not satisfy condition (2), we have
$$
N=848-34=814 \text {. }
$$

Therefore, the sum of the digits of $N$ is 13.