Specific coprime set

Let $M = p_1^{a_1} p_2^{a_2} \cdots p_n^{a_n}$ where the $p_i$ are distinct primes and the $a_i$ are positive integers. Suppose $1/N$ of the positive integers $\le M$ are coprime to $M.$ If $N=1$ then $M=1.$ Now assume $M \ge 2,$ so $N \ge 2.$

Then the fraction of integers less than or equal to $M$ which are coprime to $M$ is $\frac{(p_1-1)(p_2-1)\cdots(p_n-1)}{p_1p_2\cdots p_n}.$ If this is the reciprocal of $N,$ the power of $2$ in the numerator has to be less than or equal to the power of $2$ in the denominator. If all the $p_i$ are odd, this is impossible (unless $M=1$ ). So without loss of generality $p_1=2$ and the denominator contains exactly one $2.$ In this case, the numerator contains at most one even number, hence there is at most one other prime $p_2.$

So we get $\frac{p_2-1}{2p_2} = \frac1{N},$ whence $p_2 = \frac{N}{N-2} = 1 + \frac2{N-2},$ which is only an integer if $(N-2)|2.$ If $N=3$ we get $p_2=3$ and if $N=4$ we get $p_2=2,$ which is impossible. So $p_2=3$ is the only possibility.

Hence $S = \{1\} \cup \{2^a 3^b : a \ge 1, b \ge 0\}.$

Then $100\sum\limits_{s \in S} \frac1{s} = 100 + 100\sum\limits_{a=1}^\infty \sum\limits_{b=0}^\infty \frac1{2^a3^b} = 100 + 100 \sum_{a=1}^\infty \frac1{2^a} \sum_{b=0}^\infty \frac1{3^b} = 100 + 100 \cdot 1 \cdot 3/2 = \fbox{250}.$

For a positive integer M we get the number of coprime or Euler function Fi(n)= M(1-1/p 1)(1-1/p 2)...... Now, 1/{(1-1/p 1)(1-1/p 2)....} would be a natural number only when p 1 and p 2 are 2 and 3. So s can be of the form (2^a)(3^b). a≥1,b≥0. An 1 is also a value of s. Then the summation becomes 2.5. So, the answer comes 250.