Is there any partial sums of harmonic series that is integer? is there any partial sums of harmonic series that add up to an integer? partial sums not as trivial as the first term only i.e. 1, or the powers of 2 i.e the infinite geometric series for 2. 
This might be trivially obvious for a subset of permutations of reciprocal of some integers.
PS: there is already another solved problem that states $\sum_{k=1}^n \frac{1}{k}$ is not an integer for all $n$, but my question allows removing finite or infinite subsets of terms, excluding 1 and geometric series for 2.
 A: One way to produce partial sums that add to $1$:  Pick a natural number $N$ and write the Egyptian Fraction expansion for $\frac {N-1}N$.  Then, unless $\frac 1N$ appears in that expansion, adding $\frac 1N$ to the list gets you an example. 
For instance, starting from $N=347$ we get the example $$\frac 12 +\frac 13+\frac 17+\frac 1{48}+\frac 1{347}+\frac 1{10600}+\frac 1{154484400}=1$$  (happily, wolfram alpha knows how to compute Egyptian fractions).
This gives rise to a question which might be of some interest:  For which natural numbers $N>1$ does $\frac 1N$ appear in the Egyptian fraction expansion of $\frac {N-1}N$?  For instance $\frac 34=\frac 12+\frac 14$, or $\frac {11}{12}=\frac 12+\frac 13+\frac 1{12}$, or $\frac {83}{84}=\frac 12+\frac 13+\frac 17+\frac 1{84}$.  Further searching also turned up the examples $N=3612, 6526884$.  OEIS recognizes these as legs in integer Pythagorean triangles with some nice properties.  Should add that my search was not at all comprehensive. Once I got the pattern from OEIS I checked higher terms in that sequence and confirmed that they had the desired property. 
Edit:  just to be clear, we're referring to that expansion obtained via the greedy algorithm. Egyptian expansions aren't unique and it's easy to produce examples using "non-greedy" expansions.
A: Since $\tfrac{1}{2} + \tfrac{1}{3} < 1$, any example must involve at least $3$ fractions, and it's not hard to see that
$$\phantom{(\ast)} \qquad 1 = \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{6}, \qquad (\ast)$$
is the only example of length $3$.
One can also exploit identities among fractions of the form $\frac{1}{n}$ (called Egyptian fractions) like $$\tfrac{1}{m} = \tfrac{1}{m + 1} + \tfrac{1}{m (m + 1)} .$$ Indeed, applying this to the (inadmissible) decomposition $1 = \tfrac{1}{2} + \tfrac{1}{2}$ gives $(\ast)$, and applying this to $(\ast)$ itself gives both
$$1 = \tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{6} + \tfrac{1}{12}$$ and lulu's example,
$$1 = \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{7} + \tfrac{1}{42} .$$
There are precisely four more examples of length $4$ (for a total of $6$ such examples) and $72$ examples of length $5$. (See this calculator, which will generate all examples of a given length.)
A: You can also find infinite subseries of the harmonic series that converge to any positive real number $x$, whether $x$ is integer, rational, or irrational. The same is true for any divergent series $\sum a_n = \infty$ where $\lim_n a_n = 0$.
To find such a subseries $\sum_{i} a_{n_i} = x$, for each $j$, you choose $$n_j = \min\left\{n \in \Bbb N \mid n > n_{j-1} \text{ and } a_n < x - \sum_{j= 0}^{i - 1} a_{n_i}\right\}$$
For example,
$$\frac 1{1} + \frac 1{2}+ \frac 1{3}+ \frac 1{7}+ \frac 1{43}+ \frac 1{1807} + ... = 2$$
