19. (1985 British Mathematical Olympiad) A positive integer is called a "bad number" if the number of 1's in its binary representation is even, for example, $18=(10010)_{2}$ is a "bad number". In the set of positive integers, find the sum of the first 1985 "bad numbers".
19. We also consider 0 as a "bad number", which does not affect the result we are seeking. First, we prove the lemma using mathematical induction: In the range from 0 to $2^{n}-1 (n \geqslant 2)$, the number and sum of "bad numbers" are exactly half of the total.

(1) In the range from 0 to $2^{2}-1$, i.e., in $0,1,2,3$, 0 is a "bad number", and $3=(11)_{2}$ is a "bad number". Therefore, there are 2 "bad numbers", and $0+3=\frac{1}{2}(0+1+2+3)$. Thus, in the range from 0 to $2^{2}-1$, the number of "bad numbers" is half of the total, and their sum is also half of the total sum.

(2) Assume that in the range from 0 to $2^{n}-1$, the number and sum of "bad numbers" are exactly half of the total.
By adding $2^{n}$ to each number in the range from 0 to $2^{n}-1$, it is equivalent to adding an extra 1 in the binary representation, thus turning "bad numbers" into "non-bad numbers". Therefore, the number and sum of "bad numbers" in the range from $0+2^{n}$ to $2^{n}-1+2^{n}$, i.e., from $2^{n}$ to $2^{n+1}-1$, are also exactly half of the total. Hence, the number and sum of "bad numbers" in the range from 0 to $2^{n+1}-1$ are exactly half of the total.

This completes the proof of the lemma using mathematical induction.
Thus, the number of "bad numbers" from 0 to $2^{n}-1$ is $2^{n-1}$, and their sum is $\frac{1}{2}\left[1+2+\cdots+\left(2^{n}-1\right)\right]=2^{n-2}\left(2^{n}-1\right)$.
Since $1984=2^{10}+2^{9}+2^{8}+2^{7}+2^{6}$, the sum of the first $2^{10}$ "bad numbers" is $2^{9}\left(2^{11}-1\right)$,
The sum of the first $\left(2^{10}+2^{9}\right)$ "bad numbers" also includes the sum of "bad numbers" from $2^{11}$ to $2^{11}+2^{10}-1$, which is $2^{11} \cdot 2^{9}+2^{8}\left(2^{10}-1\right)$.
Therefore, the sum of the first $2^{10}+2^{9}+2^{8}+2^{7}+2^{6}=1984$ "bad numbers" also includes $\left(2^{11}+2^{10}\right) \cdot 2^{8}+2^{7}\left(2^{9}-1\right)+\left(2^{11}+2^{10}+2^{9}\right) \cdot 2^{7}+2^{6}\left(2^{8}-1\right)+\left(2^{11}+2^{10}+2^{9}+2^{8}\right) \cdot 2^{6}+2^{5}\left(2^{7}-1\right)$.
Thus, the sum of the first 1986 "bad numbers" (including 0, if 0 is excluded, it is the sum of the first 1985 "bad numbers") is
$$
\begin{aligned}
& 2^{9}\left(2^{11}-1\right)+2^{11} \cdot 2^{9}+2^{8}\left(2^{10}-1\right)+\left(2^{11}+2^{10}\right) \cdot 2^{8}+2^{7}\left(2^{9}-1\right)+\left(2^{11}+2^{10}+2^{9}\right) \cdot 2^{7}+ \\
& 2^{6}\left(2^{8}-1\right)+\left(2^{11}+2^{10}+2^{9}+2^{8}\right) \cdot 2^{6}+2^{5}\left(2^{7}-1\right)+\left(2^{11}+2^{10}+2^{9}+2^{8}+2^{7}\right) \cdot 2+2+1 \\
= & 2^{21}+2^{20}+2^{19}+2^{18}+2^{13}+2^{11}+2^{9}+2^{8}+2^{5}+2+1 .
\end{aligned}
$$

This is the sum of the first 1985 "bad numbers" (in decimal).