Can any partial harmonic sum be an integer? There have been a number of posts about the harmonic series, e.g. not being an integer for any number of terms.
Edit: Below I try to prove that not only H(n) but also H(2n)-H(n) is not an integer. 
I note e.g. that H(2n)-H(n) must contain a term 1/b with b prime according to Bertrand´s theorem i.e. $1/(n+1)+1/(n+2)+…+1/b+…+1/(2n) =H(2n)-H(n)$ and therefore I take it that with the least common denominator $ LCD(1/(n+1)+1/(n+2)+…+1/b+…+1/(2n))=LCD(H(2n)-H(n))$ and that
$LCD(H(2n)-H(n)) =0$ (mod b) if $H(2n)-H(n)$ is an integer. But LCD only contains b to the power of 1 and therefore LCD/b cannot be 0 whereas all the other terms – LCD/n where LCD contains b but the 1/n term does not - must be 0 (mod b), which is a contradiction.
I now wonder whether ANY sum made up of 1/n for different n - not necessarily in sequence - can be an integer? 
I cannot prove anything, but intuitively I feel that the unique combinations of the primes of any 1/n cannot be “compensated” for by any other sum of other inverted numbers to make an integer sum.
 A: It is possible:
$$1/2+1/3+1/6=1$$
In general, you might want to read up on Egyptian fractions. 
Any nonnegative rational number (in particular, any positive integer) can be expressed as a sum of distinct unit fractions. One way to do this is to proceed greedily: that is, to keep subtracting off the largest remaining unused unit fraction until your remainder is a unit fraction. For example, if we wanted to write $2$ in this way, we could notice that
$$
\sum_{i=2}^{10} \frac{1}{i} < 2 < \sum_{i=2}^{11} \frac{1}{i}
$$
so we can start by writing
$$
2 = \sum_{i=2}^{10} \frac{1}{i} + R_1
$$
for some remainder $R_1$.
Then, notice that the largest unit fraction that will "still fit" is $\frac{1}{15}$: that is,
$$
\frac{1}{15} < R_1 < \frac{1}{14}
$$
So our next approximation is
$$
2 = \sum_{i=2}^{10} \frac{1}{i} + \frac{1}{15} + R_2
$$
and, after some work, we find that $\frac{1}{230} < R_2 < \frac{1}{229}$.
So the next approximation after that is
$$
2 = \sum_{i=2}^{10} \frac{1}{i} + \frac{1}{15} + \frac{1}{230} + R_3
$$
and, this time, it turns out that $R_3$ itself is a unit fraction, $R_3=\frac{1}{57960}$. So that gives us our representation of $2$ as the sum of unit fractions:
$$
2 = \sum_{i=2}^{10} \frac{1}{i} + \frac{1}{15} + \frac{1}{230} + \frac{1}{57960}
$$
It turns out that if you do this procedure for any rational number, it always terminates (that is, eventually your remainder is always a unit fraction). The proof of this is sketched in the wikipedia article I linked to above; essentially, it boils down to showing that the numerators of the $R_i$ must decrease at each step — so there's no way this process could continue indefinitely.
On the other hand, this may not be the most efficient representation of your number as a sum of unit fractions (whether your measure of efficiency involves number of terms or size of denominators). The example Wikipedia gives is the fraction $\frac{5}{121}$, where this "greedy" method produces the sum
$$
\frac{5}{121}=\frac{1}{25}+\frac{1}{757}+\frac{1}{763309}+\frac{1}{873960180913}+\frac{1}{1527612795642093418846225}
$$
but in fact there's a much more compact representation:
$$
\frac{5}{121}=\frac{1}{33}+\frac{1}{121}+\frac{1}{363}
$$
