If we focus on the equation
$$
f(k)=\sum_{l=1}^k e^{\imath 2 \pi H(l)}
$$
for some function $H(x)$ and search for an integer $\Delta$ such that
$$
\forall k \in \mathbb{N} \qquad f(k+\Delta) = f(k),
$$
we find
\begin{eqnarray}
0
& = & \left[ f(k+1+\Delta) - f(k+1) \right] - \left[f(k+\Delta) - f(k) \right]\\
& = & \left[ f(k+1+\Delta) - f(k+\Delta) \right] - \left[f(k+1) - f(k) \right]\\
& = & e^{\imath 2 \pi H(k+1+\Delta)} - e^{\imath 2 \pi H(k+1)}
\end{eqnarray}
Hence, we find that the problem reduces to finding an integer $\Delta$ such that
$$
\forall k \in \mathbb{N} \qquad H(k+\Delta) - H(k) \in \mathbb{Z}.
$$
If we restrict ourselves to $H(x)$ being a polynomial of degree $n$ with rational coefficients, it can be formulated as:
$$
H(x) = \frac{1}{N} \sum_{i=0}^n a_i x^i,
$$
for some $N,a_i \in \mathbb{Z}$ and there is no common divisor in the set of $\{a_i\}$. It follows that
\begin{eqnarray}
H(k+\Delta) - H(k)
& = & \frac{1}{N} \sum_{i=0}^n a_i \left[(\Delta+k)^i - k^i \right]\\
& = & \frac{1}{N} \sum_{i=0}^n a_i \sum_{j=1}^i \binom{i}{j} \Delta^j k^{i-j}\\
& = & \frac{\Delta}{N} \sum_{i=0}^n a_i \sum_{j=0}^{i-1} \binom{i}{j+1} \Delta^j k^{i-j-1} \in \mathbb{Z} \qquad (*)
\end{eqnarray}
and hence that the denominator $N$ of the rational polynomial $H(x)$ is a correct solution for $\Delta$.
Note, however, that $N$ is not necessarily the smallest solution for $\Delta$, but that the smallest solution for $\Delta$ will have to be a divisor of $N$. In fact, the series of polynomial $H_n(x)=\sum_i \frac{x^i}{i}$ would have the corresponding denominators $N_n=2,6,12,60,60,\dots$ for $n \geq 2$ and the OP already established that there are smaller solutions.
Finding the smallest solution of $\Delta$ is relatively straightforward, as one can simply check the validity of $(*)$ for all $k$ by dividing out respective (prime)factors $d$ from $N$, i.e., if $N = d \Delta$ than $(*)$ requires that
$$
\sum_{i=0}^n a_i \sum_{j=0}^{i-1} \binom{i}{j+1} \Delta^j k^{i-j-1} \equiv 0 \mod d \qquad \forall k \in \mathbb{N}
$$
for which it is sufficient to check the values $0 \leq k < d$.
Addendum:
In view of the observation by the OP (25/10/2019) that the periods $\pi_n$ appear to be the primorial numbers. We consider the particular set of functions
$$
H_n(x)=\sum_{k=1}^n \frac{x^k}{k}
$$
more closely. The goal is to find the smallest $\pi_n$ such that $H_n(x+\pi_n) - H_n(x) \in \mathbb{Z}$ for all $x$. In order to do so, we focus on the simpler set of functions $f_n(x)=\frac{x^n}{n}$ and find their corresponding period. This is equivalent to solving the smallest $\pi$ for which
$$
(x+\pi)^n \equiv x^n \mod n \qquad\text{for all $x$}\qquad (**)
$$
In particular this should be true for $x=0$, and hence $n\# | \pi$, where the primorial $n\#$ is the product of all its prime factors
$$
n\# = \prod_{\text{prime } p \leq n} p
$$
In what follows we will see that $\pi=n\#$ is sufficient and therefore the smallest positive solution.
First we will show that for any prime $p$ and non-negative integer $l$, we find that for any integer $x$
$$
p^{l+1} | (x+p)^{p^l} - x^{p^l}
$$
This follows from induction by realising that for $l=0$ it says $p|(x+p)-x$ and the general factorisation expression
$$
\frac{(x+p)^{p^{(l+1)}} - x^{p^{(l+1)}}}{(x+p)^{p^l} - x^{p^l}} = (x+p)^{p^l(p-1)} + (x+p)^{p^l(p-2)}x^{p^l} + \cdots + x^{p^l(p-1)} \equiv p x^{p^l(p-1)} \equiv 0 \mod p
$$
It follows that for $n=p^l m$ with any prime factor $p$, and co-prime integer $m$ we get
$$
(x+p)^n = \left[(x+p)^{p^l}\right]^m \equiv \left[x^{p^l}\right]^m = x^n \mod p^l
$$
and therefore also any period that is a multiple of $p$ would equally be allowed, in particular
$$
(x+\pi)^n \equiv x^n \mod p^l.
$$
Since this is valid for every prime $p|n$ we have proven (**).
It therefor follows that the smallest period $\pi$ for the function $f_n(x)=\frac{x^n}{n}$ is given by
$$
\pi = n\# = \prod_{\text{prime } p|n} p
$$
As a direct consequence, we can also conclude that for $H_n(x)$ the corresponding primorial $\pi_n$ will satisfy $H_n(x+\pi_n)-H_n(x) \in \mathbb{Z}$, because it is true for each of its individual terms.
However, the constraint on the periods of all of the individual terms in $H_n(x)$ is more restrictive than the case where it only needs to hold for their combination within $H_n(x)$ itself. Hence, although the primorial $\pi_n=n\#$ is indeed a correct solution, we can not yet conclude that it is also the smallest solution for $H_n(x)$.
Best Answer
Update (As @DarkMalthorp wrote, there was a mispint)
$f(x) = \sum_{k=-n}^n |x-k|$. If we see graph of $f(x)$, then $f(0) \to \infty$. So we will use scale transform. Consider $g_n(x) =\frac{1}{n^2}{f(nx)}.$ Claim. $$g_n(x) \to g(x) =\begin{cases} 2 | x |, \ if \ |x| > 1\\ x^2 + 1, \ if \ |x| \le 1 \end{cases}, \quad n \to \infty.$$ Proof. $$ g_n(x) = \frac{1}n \sum_{k=-n}^n |x - \frac{k}n| \to \int_{-1}^1 |x - t| dt = g(x) =\begin{cases} 2 | x |, \ if \ |x| > 1\\ x^2 + 1, \ if \ |x| \le 1 \end{cases}$$
So $\frac{1}{n^2}{f(nx)} = g_n(x) \approx g(x)$ and $f(y) \approx n^2 g\big( \frac{y}n \big) $.
Addition: The idea of solution is the next one: we must choose a scaling transformation of both coordinate axes such that after it we get a function that has a finite limit
Addition2: as @Ethan Bolker mentioned, after the rescaling the finite sum is a trapezoidal approximation to the integral