Problem : Consider people of one billion, and each has one card containing one number. For instance first has card of number $7$. second has card of number $11$ and so on (simply if number means age or weight, it is fine). We want to have average and variance.
Solution : But the number is large and in reality it is impossible.
So we choose 100 people So let $x_i,\ 1\leq i\leq 100$ to be a number in card. We decide scale for instance $n=2$.
Hence $$\overline{X} = \frac{1}{2}(x_i+x_j)$$ If we allow repeatition, then we have samples $100^n = 10^4$
Hence we have from $10^4$ samples : $$E(\overline{X} )=m, \ V(\overline{X} ) = \frac{\sigma^2}{n}$$
Hence we esimate $m,\ \sigma^2$ for billion people
Question : This is right ? This is usual method ? If we consider 100 people, we have already estimation. But why do we considering $10^4$ samples ?
You have described a particular type of bootstrap procedure that is called "$m$ out of $n$ bootstrapping".
In ordinary bootstrapping, we take a data set of $n$ observations and resample from it with replacement a large number of times, obtaining new samples that are also of size $n$. For each bootstrap sample, we compute its mean, so that a histogram of bootstrap means is built. From the mean and variance of this histogram, we can then estimate the mean and variance of the underlying population using the formulas reported in the OP.
In the $m$ out of $n$ bootstrap method, we take bootstrap samples of size $m$ that are smaller than the original sample of size $n$. This alternative method is usually performed when ordinary bootstrapping fails to generate a plausible distribution and yields inconsistent results, since it has been shown that taking a smaller sample size $m$ can often lead to consistent findings. This method works asymptotically with both $m$ and $n$ tending to infinity, and with $ m/n$ tending to zero.
In your case, in particular, $n=100$, $m$ was set to $2$, and the resampling was obtained by determining all possible pairs of observations.