As stated, I tried to prove the following:
The theorem seems to be very incompletely phrased, since one can obtain non integer sums of the form
$\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{7} + \frac{1}{9}$
or
$\frac{1}{2} + \frac{1}{4} + \frac{1}{4} + \frac{1}{{18}} + \frac{1}{{20}}$
so further detail is needed. Maybe this should be closed as no longer relevant until I come up with a better phrasing and I consider more initial conditions. The big question would be
Given the set $S$ of $n$ integers
$S=\{x_1,x_2,\dots,x_n\}$ what are sufficient conditions on $x_1,x_2,\dots,x_n$ so that $\eta = \sum_{k \leq n }x_k^{-1}$ is not an integer?
Though I don't know if this is an important/relevant question to be asking.
THEOREM If $x_1,\dots,x_n $ are pairwise coprime, $x_i\neq 1$, let
$\mu =\sum_{k=1}^n \frac 1 x_k $
Then $\mu$ can't be an integer.
PROOF By induction on $n$. Asume the theorem is true for $2, \dots, n-1$. I'll analize the case $k=n$.
$(1)$ It is true for $n=2$. If $(x_1,x_2)=1 \Rightarrow (x_1 x_2,x_1+x_2)=1$
The proof is simple. We have that $(x_1,x_2)=1$. Let $d \mid x_1+x_2 , d \mid x_1x_2$. Then
$d\mid x_1(x_1+x_2)-x_1x_2 \Rightarrow d\mid x_1^2$
$d\mid x_2(x_1+x_2)-x_1x_2 \Rightarrow d\mid x_2^2$
So $d \mid (x_1^2,x_2^2)=(x_1,x_2)=1 \Rightarrow d=1$
This means $\frac{1}{x_1}+\frac{1}{x_2}=\frac{x_1+x_2}{x_1 x_2}=\phi$
is not an integer.
$(2)$ Let
$\mu = \frac{1}{x_1}+ \frac{1}{x_2}+\cdots+ \frac{1}{x_{n-1}}+ \frac{1}{x_n}$
Then
$x_n \mu-1 = x_n\left(\frac{1}{x_1}+ \frac{1}{x_2}+\cdots+ \frac{1}{x_{n-1}}\right) =x_n \omega$
By hypothesis, $(x_1,\dots,x_{n-1})=1$ so $\omega$ is not an integer. Thus, if $x_n \mu-1$ were an integer, it must be the case:
$ x_n\left(\frac{1}{x_1}+ \frac{1}{x_2}+\cdots+ \frac{1}{x_{n-1}}\right) =k \text{ ; } k \text{ an integer }$
$ x_n \frac{\tau}{x_1 x_2 \cdots x_{n-1}} =k \text{ ; } k \text{ an integer }$
$\tau$ is the numerator obtained upon taking a common denominator.
But since $\omega$ is not an integer, then it must be the case
$x_1 x_2 \cdots x_{n-1} \mid x_n$
which is imposible. Then $x_n \mu -1$ is not an integer. But since $x_n$ and $1$ are, this means $\mu$ isn't an integer, this is,
$\mu =\sum_{k=1}^n \frac 1 x_k $
is not an integer. $\blacktriangle$
NOTE The hypothesis that $x_k \neq 1$ is necessary to avoid sums like
$\frac{1}{1}+\overbrace{\frac{1}{n}+\cdots +\frac{1}{n}}^{n }=1+n\frac{1}{n}=2$
however, if $(x_1,\dots,x_n)=1$, the sum
$\nu =\sum_{k=1}^n \frac 1 x_k +1 $
will clearly not be an integer.