Fix an integer $n$ ≥ 2. For which real numbers $x$ is $\lfloor nx\rfloor$ - ${\sum }_{k=1}^n$ (⌊kx⌋ / k) maximal, and what is the maximal value that this expression can take?

Step 1 (Reduction to fractional part): Let $x=\lfloor x\rfloor +\{x\}$. Since $\lfloor kx\rfloor =k\lfloor x\rfloor +\lfloor k\{x\}\rfloor$, the expression simplifies to:

Thus, we can assume without loss of generality that $x\in [0,1)$.

Step 2 (Upper Bounding): Since $x<1$, we have $\lfloor kx\rfloor \le k-1$ for all $1\le k\le n$. Therefore, each term $\frac{\lfloor kx\rfloor }{k}$ is at most $\frac{k-1}{k}=1-\frac{1}{k}$. To maximize the overall expression, we want to choose $x$ close to $1$ such that all $\lfloor kx\rfloor$ achieve their maximum value of $k-1$.

Step 3 (Finding the Optimal Interval): Let us choose $x$ in the interval $[1-1/n,1)$. For any $k\in \{1,2,\dots ,n\}$:

Since $k\le n$, we have $\frac{k}{n}\le 1$, meaning $kx\ge k-1$. Since $x<1$, we also have $kx<k$. Therefore, for all $k\in \{1,2,\dots ,n\}$, we have exactly:

Substituting this into our expression yields:

Since this achieves the upper bound of individual differences, the maximal value is exactly $H_n-1$. $■$