In one of T. Apostol's student textbooks on analytic number theory (i.e., Introduction to Analytic Number Theory, T. Apostol, Springer, 1976) Wolstenholme's theorem is stated (Apostol, Chapt. 5, page 116) as follows (more or less):
For any prime ($p \geq 5$),
\begin{equation} ((p - 1)!)(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p - 1}) \equiv 0 \pmod {p^2}. \end{equation}
Suppose one was to multiply through both sides of the congruence with the inverse of $(p - 1)!$ modulo $p^{2}$. One gets
\begin{equation} 1\cdot(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{p - 1}) \equiv 0 \pmod {p^2}. \end{equation}
Does this congruence "make sense," since one has a finite sum of fractions on the left hand side, not a finite sum of integers (Cf. Apostol, Exercise 11, page 127)?
By the expression ``inverse of $(p - 1)!$ modulo $p^{2}$," I mean multiplying through by $t$, so that
$$((p - 1)!)t \equiv 1 \pmod{p^2}.$$
Just asking. After all I do not know everything.... :-)
In modular arithmetic, you should interpret Egyptian fractions (of the form $\frac 1a$) as the modular inverse of $a \bmod p^2$, in which case this makes perfect sense.