Subsequence and integers as a sum of $\frac{1}{n}$ For all $M \in \mathbb{Z}$, is there a finite sequence of positive integers (not necessarily distinct) $(n_i)_{i \in I}$, s.t. $\sum_{i \in I} \frac{1}{n_i} = M$, and there is no subsequence $(n_i)_{i \in J}$ of $(n_i)_{i \in I}$ such that $\sum_{i \in J} \frac{1}{n_i}$ is an integer ?
 A: You have answered your question in the comments, just construct your sequence inductively.
Suppose you have a sequence $n_1,n_2,\ldots,n_k$ of pairwise relatively prime numbers. Put $q=n_1n_2\ldots n_k$. You put $q_i={q\over n_i}$ and find $d_i<n_i$ such that $d_iq_i+1=0$ mod $n_i$. This implies that $d_1q_1+d_2q_2+\ldots+d_nq_n+1=0\mathop{\rm mod}n_i$ for $i=1,2,\ldots,k$ hence $\mathop{\rm mod}q$. Dividing by $q$ we get some integer $$M={d_1\over n_1}+{d_2\over n_2}+\ldots+{d_k\over n_k}+{1\over q}$$.
So far this was your argument.
Now pick any $n_{k+1}$, $1<n_{k+1}<q$, relatively prime to every $n_i$ for $i\leq k$. Find $n_{k+2}<q$ such that $n_{k+1}n_{k+2}=1\mathop{\rm mod}q$. The numbers $n_{k+1}$ and $n_{k+2}$ may have a common factor but you may prevent that by choosing $n_{k+1}$ to be a prime bigger than $q/2$ (Edit: and such that $n_{k+1}^2\neq 1\mathop{\rm mod} q$, for example of the form $5r\pm 2$ if $5|q$, see comments by pallab1234 below). Put $q'=qn_{k+1}n_{k+2}=n_1n_2\ldots n_{k+2}$. When you find $d_i<n_i$ such that $d_iq'_i+1=0$ mod $n_i$ for $i=1,2,\ldots,{k+2}$ you see that those for $i\leq k$ didn't change. You obtain another integer 
$$
M'={d_1\over n_1}+\ldots+
{d_k\over n_k}+{d_{k+1}\over n_{k+1}}+{d_{k+2}\over n_{k+2}}+{1\over q'}=
M+\Delta
$$
where
$$
\Delta={d_{k+1}\over n_{k+1}}+{d_{k+2}\over n_{k+2}}+{1\over q'}-{1\over q}=
{d_{k+1}\over n_{k+1}}+{d_{k+2}\over n_{k+2}}+
{1\over q}\left({1\over n_{k+1}n_{k+2}}-1\right)
$$
must be an integer. Clearly $\Delta<2$. On the other hand
$$
\Delta>{1\over n_{k+1}}+{1\over n_{k+2}}-{1\over q}>{1\over n_{k+2}}-{1\over q}
$$
which is positive since $n_{k+2}<q$. Being an integer $\Delta=1$ hence $M'=M+1$.
Edit: deleted.
Edit2: Thus you may start with $n_1=3$ and $n_2=5$ for which you get $\displaystyle M={1\over 3}+{3\over 5}+{1\over 3\cdot 5}=1$ and go up to any positive integer you want.
