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 ?



Subsequence and integers as a sum of $\frac{1}{n}$
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.


