$11 \cdot 20$ A positive integer is called a "bad number" if the number of 1s in its binary representation is even, for example, $18=(10010)_{2}$ is a "bad number".
Among the set of positive integers, find the sum of the first 1985 "bad numbers".
(British Mathematical Olympiad, 1985)
【Solution】We consider 0 as a "bad number" as well, which does not affect the result we are looking for.

First, we use mathematical induction to prove the lemma:
In the range from 0 to $2^{n}-1 (n \geqslant 2)$, the number and sum of "bad numbers" are exactly half of them.
(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", so there are 2 "bad numbers". And
$$
0+3=\frac{1}{2}(0+1+2+3) .
$$

Therefore, in the range from 0 to $2^{2}-1$, the number of "bad numbers" is half of them, and their sum is also half of their 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 them.

Since adding $2^{n}$ to each number from 0 to $2^{n}-1$ is equivalent to adding an extra 1 in the binary representation, the "bad numbers" from 0 to $2^{n}-1$ all become "non-bad numbers", thus the number and sum of "bad numbers" 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 them. Therefore, the number and sum of "bad numbers" from 0 to $2^{n+1}-1$ are exactly half of them.
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
$$
2^{11} \cdot 2^{9}+2^{8}\left(2^{10}-1\right) \text {. }
$$

Therefore, the sum of the first $2^{10}+2^{9}+2^{8}+2^{7}+2^{6}=1984$ "bad numbers" also includes
$$
\begin{array}{l}
\left(2^{11}+2^{10}\right) \cdot 2^{8}+2^{7}\left(2^{9}-1\right)+\left(2^{11}+2^{10}+2^{9}\right) \cdot 2^{7} \\
+2^{6}\left(2^{8}-1\right)+\left(2^{11}+2^{10}+2^{9}+2^{8}\right) \cdot 2^{6}+2^{5}\left(2^{7}-1\right) .
\end{array}
$$

Thus, the sum of the first 1986 "bad numbers" (including 0) (if not including 0, it is the sum of the first 1985 "bad numbers") is
$$
\begin{array}{l}
2^{9}\left(2^{11}-1\right)+2^{11} \cdot 2^{9}+2^{8}\left(2^{10}-1\right)+\left(2^{11}+2^{10}\right) \cdot 2^{8} \\
+2^{7}\left(2^{9}-1\right)+\left(2^{11}+2^{10}+2^{9}\right) \cdot 2^{7}+2^{6}\left(2^{8}-1\right) \\
+\left(2^{11}+2^{10}+2^{9}+2^{8}\right) \cdot 2^{6}+2^{5}\left(2^{7}-1\right)
\end{array}
$$
$$
\begin{array}{l}
+\left(2^{11}+2^{10}+2^{9}+2^{8}+2^{7}\right) \cdot 2+2+1 \\
=2^{21}+2^{20}+2^{19}+2^{18}+2^{13}+2^{11}+2^{9}+2^{8}+2^{5}+2+1 .
\end{array}
$$

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