$$H_{\frac{p-1}{2}} \equiv -2\cdot q_{p,2} \pmod p $$ $$H_{\frac{p-1}{2}} \equiv -2\cdot q_{p,2}+p\cdot q_{p,2}^2\pmod{p^2}$$
$p$ is an odd prime.
$H_n=\sum_{1\le j \le n}\frac{1}{j}$ is a harmonic number,
$q_{p,2}=\frac{2^{p-1}-1}{p}$ is the Fermat quotient with base $2$.
A quite easy proof of the first congruence goes like this :
First, one writes $\binom{p-1}{j}=\frac{1}{j!}(p-1)(p-2)\cdot \cdot (p-j)$ . By doing the development and keeping the two lowest powers of $p$ only, this gives: $$\binom{p-1}{j} \equiv (-1)^j +(-1)^{j-1}p\cdot H_j \pmod {p^2}$$ Then: $$q_{p,2}=\frac{2^{p-1}-1}{p}=\frac{1}{p}\sum_{1\le j \le {p-1}}\binom{p-1}{j}$$ $$p\cdot q_{p,2}\equiv \sum_{1\le j \le {p-1}}\left((-1)^j +(-1)^{j-1}p\cdot H_j\right) \pmod {p^2}$$ Hence$$q_{p,2}\equiv\sum_{1\le j \le {p-1}}(-1)^{j-1} H_j \pmod {p}$$ Then pairing the consecutive summands on the RHS $$q_{p,2}\equiv -\sum_{1\le j \le \frac{p-1}{2}}( H_{2j}-H_{2j-1}) \pmod {p}$$ $$q_{p,2}\equiv -\sum_{1\le j \le \frac{p-1}{2}}\frac{1}{2j} \pmod {p}$$ $$2\cdot q_{p,2}\equiv -H_{\frac{p-1}{2}} \pmod {p} \ \ \ \square $$
A proof of the second congruence was given in E. Lehmer, On congruences involving Bernoulli numbers and quotients of Fermat and Wilson, Ann. Math. 39 (1938) 350–360. A more recent one can be found in this paper. But these proofs are quite involved: they are obtained as sort of corollaries amongst many others complicated and more advanced congruences between Harmonic numbers and Fermat quotients, invoking Bernoulli polynomials, Bernoulli numbers and the Staudt-Von Clausen theorem.
So my question is: Can someone help me finding a more elementary proof for the second congruence above, that would be as similar as possible to the first one, i.e. without Bernoulli things, but going only with tricks, pairing or rearranging the summands, using properties of binomial coefficients or playing with symetries etc. ?