That's no way to behave

$S$ is the set of all "impolite" numbers, (hence the title of the question), which is composed of all non-negative integer powers of $2.$ Thus the sum of the elements of $R$ is

$\displaystyle\sum_{k=0}^{\infty} \dfrac{1}{2^{k}} = \boxed{2}.$

I'll provide a proof later regarding the composition of $S$ , but for now I just wanted to get something posted in case anyone has questions/concerns. Here is a link regarding polite/impolite numbers.

We can prove that $S$ is precisely the set of powers $2^i$ where $i=0,1,\ldots$ .

Part 1: $1$ belongs to $S$

This is clear from the definition of $S$ - but is a special case.

Part 2: if $n$ can be written as the sum of $k$ positive integers, it must have an odd factor.

Suppose we have $n=a+(a+1)+(a+2)+\cdots+(a+k-1)$ . This can summed to give $n=\frac{k(2a+k-1)}{2}$ .

Exactly one of the numbers $k$ and $2a+k-1$ is even and the other is odd; so $n$ always has an odd factor.

This tells us that all powers of $2$ greater than $1$ belong to $S$ .

Part 3: if $n$ has an odd factor greater than $1$ , it can be written as a sum of consecutive positive integers.

Say $d>1$ is an odd factor of $n$ . Then the $d$ consecutive integers centred around the integer $\frac{n}{d}$ total to $n$ . For example, take $n=30$ and $d=5$ ; this method gives the sum $4+5+6+7+8=30$ .

However, this runs into a problem for large $d$ and small $n$ : take $n=14$ and $d=7$ ; the sum we get is $-1+0+1+2+3+4+5=14$ . This is obviously true, but we have to find $n$ as a sum of positive integers.

We can easily get round this though: we can just cancel any negative integers in the string with their positive counterparts and leave a valid sum. In the example above, we remove $-1,0,1$ from the sum to leave $2+3+4+5=14$ . Since the string of consecutive integers is centred on a positive integer, there are always more positive terms than non-positive ones; and since there are an odd number of terms in the sum, and we always remove an odd number of terms from it (some negatives, their counterparts, and zero), we are always left with at least two terms in the final sum.

One more example for this: $33=\color{#D61F06} {-2+(-1)+0+1+2} \color{#333333} +3+4+5+6+7+8=3+4+5+6+7+8$

This part (and part 1) tells us that only powers of $2$ (including $1$ ) belong to $S$ .

Finally, the sum of reciprocals of the members of $S$ is $\boxed2$ .

Suppose $N=2^{a_0}*(p_1)^{a_1}*(p_2)^{a_2}*...$ (where $p_n$ is a prime and $a_n$ is an integer for all $n$ .

Assume $a_k$ exists as the first non-zero power (excluding $a_0$ . If $(p_k)^2-(p_k)< 2N, N = \frac{2N-(p_k)^2+(p_k)}{2p_k}+(\frac{2N-(p_k)^2+(p_k)}{2p_k}+1) + ... + \frac{2N+(p_k)^2-(p_k)}{2p_k}$ is a sum of positive integers.

If $(p_k)^2-(p_k) >= 2N$ , $(p_k)-1 >= 2^{a_0 + 1}*(p_k)^{a_k-1}*(p_{k+1})^{a_{k+1}}*....$ Hence $a_k = 1$ and $a_n = 0$ for all $n \neq 0,k$ , and $2^{a_0 + 1} < (p_k)$ .

$N = \frac{p_k+1-2^{a_0+1}}{2}+(\frac{p_k+1-2^{a_0+1}}{2}+1)+...+\frac{p_k-1+2^{a_0+1}}{2}$ is a sum of positive integers.

If no such $a_k$ exists (eg $N$ is a power of $2$ ), the only sum of consecutive integers equalling $2^{a_0}$ is $2^{a_0}= 1-2^{a_0} + 2-2^{a_0} +...+ 0 +...+ 2^{a_0}-1 + 2^{a_0}$ where, $1-2^{a_0}$ is not a positive integer.

Hence, $S$ is the set of natural powers of $2$ , and the sum of the reciprocals is $2$ .