$11 \cdot 8$ Let $n$ be a fixed positive integer. Find the sum of all positive integers that satisfy the following property: in their binary representation, they are exactly $2n$ digits long, with $n$ ones and $n$ zeros, but the first digit is not 0.
(23rd Canadian Mathematics Competition, 1991)
[Solution] (1) When $n=1$, by the problem's condition, we only have $(10)_2=2$.
That is, the sum we are looking for is 2.
(2) When $n \geqslant 2$, the first digit is 1, then the remaining $2 n-1$ positions have $n-1$ positions for 1 and $n$ positions for 0, so there are $C_{2 n-1}^{n-1}$ possibilities.

Considering any position other than the first one, the number of numbers with a digit 1 is $C_{2 n-2}^{n-2}$, and the number of numbers with a digit 0 is $C_{2 n-2}^{n-1}$. Therefore, the sum of the digits in this position is
$$
C_{2 n-2}^{n-2}=C_{2 n-2}^{n} \text{. }
$$

Thus, the sum of all numbers is
$$
\begin{aligned}
& C_{2 n-2}^{n}\left(1+2+\cdots+2^{2 n-2}\right)+C_{2 n-1}^{n-1} 2^{2 n-1} \\
= & C_{2 n-2}^{n}\left(2^{2 n-1}-1\right)+C_{2 n-1}^{n} 2^{2 n-1} .
\end{aligned}
$$