Consider the Bernoulli numbers denoted by $B_n$, which are rational numbers.
It is known that the harmonic numbers $H_n=\sum_{k=1}^n\frac1k$ are not integers once $n>1$.
I am curious about the following:
Question: If $n>0$, will $\sum_{k=0}^nB_k$ ever be an integer?
It can never be an integer for $n>0$. There is a result by K.G.C. von Staudt and independently by T. Clausen that $$B_n+\sum_{p\in \mathbb{P}\, ,\, p-1|n}\frac{1}{p}\in \mathbb Z$$
[1] T. Clausen. Lehrsatz aus einer Abhandlung uber die Bernoullischen Zahlen. Astr. Nachr., 17:351–352, 1840
[2] K. G. C. von Staudt. Beweis eines Lehrsatzes die Bernoulli'schen Zahlen betreffend. J. Reine Angew. Math., 21:372–374, 1840
Using this we see that the integrality of $\sum_{k=0}^n B_k$ is equivalent to the integrality of $$\sum_{k=2}^n\left(\sum_{p\in \mathbb P \, , \, p-1|k}\frac{1}{p}\right)=\sum_{p\in \mathbb P}\frac{\lfloor\frac{n}{p-1}\rfloor}{p}$$ If $q$ is the largest prime $\le n$ we have $\lfloor \frac{n}{q-1}\rfloor =1$ therefore our expression has $q$-valuation $-1$, so is not an integer.
It is never an integer. By Bertrand postulate there exists a prime $p=2s+1$ between $n/2$ and $n$, it divides the denominator of $B_{p-1}$ and not of other summands in your sum, by von Staudt - Clausen Theorem.
