It is true that, for $n$ large enough, $p_n(x)$ will have distinct real roots, which I will prove by induction on $m$.
Before proceding with the proof, let me first state what it shows about the behaviour of $p_n$ as $n$ increases. Assuming the leading coefficient is positive, $p_n^{(m-2)}$ are quadratics with a unique minimum. As $n$ increases, the minimum decreases monotonically until it becomes negative, at which point $p_n^{(m-2)}$ has distinct real roots. Then, $p_n^{(m-3)}$ is a cubic with a local maximum and a local minimum. As $n$ increases further, the local maximum increases until it is positive and the local minimum decreases until it is negative, at which point $p_n^{(m-3)}$ has distinct real roots. More generally, for each $k=0,1,2,\ldots,m-2$, once $p_n^{(k+1)}$ has distinct real roots then $p_n^{(k)}$ has $m-k-1$ local extrema. Further increasing $n$, the local maxima increase monotonically until they are positive and the local minima decrease until they are negative, at which point $p_n^{(k)}$ has distinct real roots. By induction then, $p_n$ eventually has distinct real roots.
Now, to continue with the proof.
For $m=1$, $p_n(x)$ is of first order so always has a real root.
For $m > 1$, we now suppose that the statement holds for polynomials of degree $m-1$. Then, setting $q_n(x)=p_n^\prime(x)$, we have $q_n(x)=q_{n-1}(x)+q_{n-1}^\prime(x)$. By the induction hypothesis, $q_n(x)$ has $m-1$ distinct real roots for all large enough $n$. I'll denote these by $a_{n,1} > a_{n,2} > \cdots > a_{n,m-1}$. These are the turning points of $p_n(x)$. I'll show that the for large $n$, the signs of $p_n(a_{n,1}),p_n(a_{n,2}),p_n(a_{n,3}),\ldots$ are alternating, from which it follows that $p_n$ has $m$ distinct roots.
By scaling, we assume wlog that $p_n(x)$ has leading coefficient $1$ (i.e., it is monic). Then, $q_n$ is positive on $(a_{n,1},\infty)$, negative on $(a_{n,2},a_{n,1})$, positive on $(a_{n,3},a_{n,2}))$, etc.
Note that $a_{n,i}$ is a local maximum of $(-1)^ip_n(x)$ (for each $i=1,2,\ldots,m-1$). This means that
$$
(-1)^ip_n(a_{n,i}+h)=(-1)^ip_n(a_{n,i})-ch^{2r}+O(h^{2r+1})
$$
for some positive $c$ and positive integer $r$. So,
$$
(-1)^ip_{n+1}^\prime(a_{n,i}+h)=-c(2r)(2r-1)h^{2(r-1)}+O(h^{2r-1}).
$$
This shows that $(-1)^ip_{n+1}(x)$ is decreasing in a neighbourhood of $a_{n,i}$.
Now, for $i=1,2,\ldots,m-2$, we see that $(-1)^ip_{n+1}(x)$ is increasing at $a_{n,i+1}$ and decreasing at $a_{n,i}$. Hence, it has a local maximum in $(a_{n,i+1},a_{n,i})$, which we will denote by $b_{i}$, so $a_{n,i+1} < b_i < a_{n,i}$ and $(-1)^ip_{n+1}(b_i) > (-1)^ip_{n+1}(a_{n-i})$. Similarly, $(-1)^{m-1}p_{n+1}(x)$ is decreasing at $a_{n,m-1}$ and (as $p_{n+1}$ is monic of degree $m$), it tends to $-\infty$ as $x\to-\infty$. Hence, it has a local maximum, $b_{m-1}$ in the range $(-\infty,a_{n,m-1})$. So, $b_{m-1} < a_{n,m-1}$ and $(-1)^ip_{n+1}(b_{m-1}) > (-1)^ip_{n+1}(a_{n,m-1})$. As $b_1 > b_2 > \cdots > b_{m-1}$ are roots of $p_{n+1}^\prime(x)$, we have $a_{n+1,i}=b_i$.
So far, we have shown that
$$
a_{n,1} > a_{n+1,1} > a_{n,2} > a_{n+1,2} > a_{n,3} > \cdots > a_{n,m-1} > a_{n+1,m-1}
$$
and, for each $i$, $(-1)^ip(a_{n,i})$ is strictly increasing in $n$. If, for each fixed $i$, it can be shown that there is an $\epsilon > 0$ with $(-1)^ip_{n+1}(a_{n+1,i})\ge(-1)^ip_n(a_{n,i})+\epsilon$ for all large $n$ then, for $n$ large enough, $(-1)^ip_n(a_{n,i})$ will be positive. So, the signs of $p_n(a_{n,1}),p_n(a_{n,2}),\ldots$ will be alternating in sign and we are done. Let us proceed by contradiction and assume to the contrary that $(-1)^ip_{n+1}(a_{n+1,i})< (-1)^ip_n(a_{n,i})+\epsilon$ (for small enough $\epsilon$ this will give a contradiction regardless of $n$).
If $i < m-1$ then $(-1)^ip_{n+1}(x)\le(-1)^ip_{n+1}(a_{n+1,i})$ on $(a_{n,i+1},a_{n,i})$. Setting $a_{n,m}=-\infty$ then this also holds for $i=m-1$. By the assumption, $(-1)^ip_{n+1}(x)\le(-1)^ip_{n}(a_{n,i})+\epsilon$ in this range so, setting $f(x)=(-1)^i(p_n(a_{n,i})-p_{n}(x))\ge0$,
$$
f(x)+f^\prime(x)=(-1)^i(p_n(a_{n,i})-p_{n+1}(x))\ge-\epsilon.
$$
Therefore,
$$
f(x)=-\int_x^{a_{n,i}}\frac{d}{dy}(e^{y-x}f(y))dy
=-\int_x^{a_{n,i}}e^{y-x}(f(y)+f^\prime(y))dy\le\epsilon\int_x^{a_{n,i}}e^{y-x}dy=\epsilon(e^{a_{n,i}-x}-1).
$$
This shows that in the range $(a_{n,i+1},a_{n,i})$,
\begin{align}
\lvert p_{n}(x)-p_n(a_{n,i})\rvert\le\epsilon(e^{a_{n,i}-x}-1).&&{\rm(1)}
\end{align}
So, $p_n$ is almost constant here for $\epsilon$ small. I'll need the following lemma, which I'll prove in a moment.
Lemma: Let $p$ be a monic polynominal of degree $m\ge1$. Then, for each $a < b$ we have
$$
sup_{x\in(a,b)}\lvert p(x)-p(b)\rvert\ge L(b-a)^m.
$$
where $L$ is a constant depending only on $m$.
We can apply this to (1). For the case $i=m-1$ and any $K>0$, apply to the range $[a_{n,m-1}-K,a_{n,m-1}]$ to get
$$
\epsilon(e^K-1)\ge L K^{m}
$$
which gives a contradiction if $\epsilon$ is small enough.
On the other hand, for $i< m-1$, $(-1)^i(p_n(a_{n,i})-p_n(a_{n,i+1})$ is positive and increasing in $n$ for all large enough $n$, so is greater than some fixed $\delta$. Applying (1),
$$
\delta\le (-1)^i(p_{n,i}(a_{n,i})-p_{n,i}(a_{n,i+1}))\le\epsilon(e^{a_{n,i}-a_{n,i+1}}-1).
$$
Assuming that $\epsilon \le 1$, we choose $K > 0$ such that $\delta\ge e^K-1$. Then, $K\le a_{n,i}-a_{n,i+1}$ and we can apply the lemma to the range $[a_{n,i}-K,a_{n,i}]$ to get
$$
\epsilon(e^K-1)\ge LK^m
$$
again, giving a contradiction for small $\epsilon$. QED
I'll now prove the lemma. For any $\epsilon\in\mathbb{R}$ define the linear operator $T_\epsilon$ of the functions $f\colon\mathbb{R}\to\mathbb{R}$ by $T_\epsilon f(x)=f(x+\epsilon)$. If $f$ is a polynomial of degree $m$ with leading coefficient $c$, then $T_\epsilon f-f$ is a polynomial of degree $m-1$ with leading coefficient $mc\epsilon$. Hence, $(T_\epsilon-1)^mf(x)=m!c\epsilon^m$. On the other hand $\lVert T_\epsilon\rVert=1$ and $(T_\epsilon-1)^mf(x)$ only depends on the values of $f(y)$ with $y$ in the range $[x,x+m\epsilon]$. So, for a monic polynomial $f$ of degree $m$,
$$
m!\epsilon^m=(T_\epsilon-1)^mf(x)\le 2^m\sup_{y\in[x,x+m\epsilon]}\lvert f(y)\rvert
$$
If $p$ is a monic polynomial of degree $m$ and $a < b$, then taking $f(y)=p(y)-p(b)$, $x=a$ and $\epsilon=(b-a)/m$ gives the lemma with $L=m!(2m)^{-m}$.
Best Answer
Here is the solution in the comments, typed out as an answer.
The trick is to separate out the negative terms:
$$1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdots - \frac{1}{1318} + \frac{1}{1319}$$ $$= 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{1319} - 2 \left( \frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{1318} \right)$$ $$=\frac{1}{660} + \frac{1}{661} + \cdots + \frac{1}{1319}.$$
and to notice that $660 + 1319 = 1979$. Combine terms in pairs from the outside:
$$\frac{1}{660} + \frac{1}{1319} = \frac{1979}{660 \cdot 1319}$$ $$\frac{1}{661} + \frac{1}{1318} = \frac{1979}{661 \cdot 1318}$$ $$\cdots =\cdots$$
and so on.
There are an even number of terms, so this gives us a sum of terms in the form $\frac{1979}{m}$, with m not divisible by $1979$ (since $1979$ is prime and so does not divide any product of smaller numbers).
Hence the sum of the terms gives a rational number with denominator not divisible by $1979$. Q.E.D