I tracked down Nagell's paper in early 1992, since I had found a proof and wanted to see whether it was the same as his. (It turned out to be essentially the same idea.) Unfortunately, I've since lost my photocopy of his paper, but it came from UIUC, so that would be where I'd start looking. If I remember right, the journal where it appeared is incredibly obscure, and UIUC was the only place in North America that had a copy.

Here's a proof copied from my very old TeX file. A bit awkwardly written, but it explains how to do the case analysis, which is the part that makes this approach simpler than what Erdős did.

Suppose that $a$, $b$, and $n$ are positive integers and
$$\frac{1}{a}+\frac{1}{a+b}+\frac{1}{a+2b}+\cdots+\frac{1}{a+nb} = c \in
\hbox{\bf Z}$$ We can take $\gcd(a,b) = 1$ and $a > 1$, and it is easy to
show that $n > 2$.

Suppose that $b$ is odd. In the arithmetic progression, there is then a
unique number $a + mb$ divisible by the highest possible power of $2$,
because for all $k$, the progression runs through a cycle modulo $2^k$ which
contains each value exactly once. Then multiplication by $\ell =
\hbox{lcm}(a,a+b,\ldots,a+nb)$ gives
$$\frac{\ell}{a}+\frac{\ell}{a+b}+\frac{\ell}{a+2b}+\cdots+\frac{\ell}{a+nb}
= \ell c.$$ All of the terms here except $\frac{\ell}{a+mb}$ are even.
Thus, $b$ cannot be odd.

Now suppose that $b$ is even, and $b \le \frac{n-2}{3}$. By Bertrand's
Postulate, there is a prime $p$ such that $\frac{n+1}{2} < p < n+1$. Then
$p$ does not divide $b$, and $p$ is odd. We must have $a \le n$, because
$$1 \le c = \frac{1}{a}+\frac{1}{a+b}+\cdots+\frac{1}{a+nb} < \frac{n+1}{a}.$$

Since $b$ generates the additive group modulo $p$ and $n+1 > p$, at least
one of the numbers $a$, $a+b$, \ldots, $a+nb$ is divisible by $p$. At most
two are, since $2p > n+1$. Suppose that $p$ divides only the term $a+kb$.
Then
$$\frac{1}{a+kb} = c-\sum_{j \neq k}{\frac{1}{a+jb}}.$$
The denominator of the left side is divisible by $p$, but that is not true
of the right side. Thus, $p$ must divide two terms.

Now suppose that $p$ divides $a+{\ell}b$ and $a+kb$, with $mp = a+{\ell}b <
a+kb = (m+b)p$. Then
$$\frac{1}{p}\left(\frac{2m+b}{m(m+b)}\right) =
c - \sum_{j \neq \ell,k}{\frac{1}{a+jb}}.$$
This implies that $p \mid (2m+b)$. However,
$$a+{\ell}b \le n + (n-p)\left(\frac{n-2}{3}\right) < n + \frac{n}{2}\left(\frac{n-2}{3}\right).$$
Therefore,
$$m < \frac{a+{\ell}b}{n/2} < \frac{n-2}{3} + 2.$$
It follows that $2m+b \le n+1$. However, $n+1 < 2p$, so $2m+b = p$. This
contradicts the fact that $b$ is even and $p$ is odd.

Finally, suppose that $b$ is even, and $b \ge \frac{n-1}{3}$. Since $a > 1$
and $a$ is odd, we must have $a \ge 3$. We must also have $b \ge 4$, since
if $b=2$, then Bertrand's Postulate and the fact that $n \ge a$ (as above)
imply that one of the terms is a prime, which does not divide any other
term.

Now, we show that $n \le 13$. To do that, note that
$$\frac{1}{a}+\frac{1}{a+b}+\cdots+\frac{1}{a+nb} < \frac{1}{3} + \frac{1}{7} + \frac{3}{n-1}\left(\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right).$$
A simple computation shows that for $n \ge 14$,
$$\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} < \frac{11}{63}(n-1),$$
which implies that the sum is less than 1. Hence, we must have $n \le 13$.

Thus, the sum is at most
$$\frac{1}{3} + \frac{1}{7} + \frac{1}{11} + \cdots + \frac{1}{47} + \frac{1}{51} + \frac{1}{55} < 1.$$

Therefore, the sum is never an integer, and the theorem holds.