For which positive integers $n$ is the sum $\sum_{k=1}^n 1 / \varphi(k)$ an integer? Here $\varphi$ is the Euler totient function.
The question is a "totient-analog" of the well-known result that $\sum_{k=1}^n 1/k$ is an integer only for $n=1$, which can be proved either using Bertrand's postulate or considering the $2$-adic valuation.
Here is some computational evidence that $n\in\{1,2,4\}$ are the only $n$ that deliver an integer sum.
For an odd prime $p$, let $a_p < b_p$ denote two smallest even positive integers such that $a_pp+1$ and $b_pp+1$ are prime. Then $n\in \big[a_pp+1,\min\{b_pp+1,p^2\}\big)$ cannot deliver an integer sum as its denominator will necessarily contain $p$. Here are these intervals for primes $p<100$:
3 [7, 9)
5 [11, 25)
7 [29, 43)
11 [23, 67)
13 [53, 79)
17 [103, 137)
19 [191, 229)
23 [47, 139)
29 [59, 233)
31 [311, 373)
37 [149, 223)
41 [83, 739)
43 [173, 431)
47 [283, 659)
53 [107, 743)
59 [709, 827)
61 [367, 733)
67 [269, 1609)
71 [569, 853)
73 [293, 439)
79 [317, 1423)
83 [167, 499)
89 [179, 1069)
97 [389, 971)
As we can see primes 5, 11, 29, 41, 67 alone cover the interval $[11, 1609)$, while others provide extensive backup support. Such a pattern will likely continue to cover all integers above $11$. However I'm not sure whether this observation can be justified theoretically without relying on unproved conjectures.
ADDED. Primes 5, 11, 29, 41, 67, 743, 7823, 165293, 6215171 cover the interval $[11,1131161123)$.
