It is well-known (Wolstenholme Theorem) that for any prime $p$ such that $p>3$ the Harmonic number $H_{p-1}$ satisfies the congruence $$ H_{p-1}:= \sum_{i=1}^{p-1}\frac{1}{i}\equiv 0 \pmod {p^2}$$ Now, for $p>5$, it seems to hold that
$$ \sum_{i=1}^{p-3}\frac{1}{p-2-i} \sum_{j=1}^i \frac{H_j}{j+1}\equiv 0 \pmod {p^2}$$
ex for $p=7$ $$\frac{\frac{1}{2}}{4} +\frac{\frac{1}{2}+\frac{1+\frac{1}{2}}{3}}{3} +\frac{\frac{1}{2}+\frac{1+\frac{1}{2}}{3}+\frac{1+\frac{1}{2}+\frac{1}{3}}{4}}{2}+\frac{\frac{1}{2}+\frac{1+\frac{1}{2}}{3}+\frac{1+\frac{1}{2}+\frac{1}{3}}{4}+\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{5}}{1}=\frac{49}{16}$$
This is probably known, but I could not find a reference.
By manipulating the sums, I can prove that for any natural number $n$ such that $n>3$ we have $$\sum_{i=1}^{n-3}\frac{1}{n-2-i} \sum_{j=1}^i \frac{H_j}{j+1} =\frac{n}{8}\sum_{i_1+i_2+i_3+i_4=n}\frac{1}{i_1i_2i_3i_4}$$ where the sum on rhs is over all the $4$-uples of positive integers whose sum is $n$. But how can we factor $n$ out of $\sum_{i_1+i_2+i_3+i_4=n}\frac{1}{i_1i_2i_3i_4}$, when $n>5$?
Edit Here are the manipulations that I have done : \begin{align*} S:=&\sum_{i=1}^{n}\frac{1}{n+1-i} \sum_{j=1}^i \frac{H_j}{j+1}\\ =&\sum_{i=0}^{n-1}\frac{1}{i+1} \sum_{j=1}^{n-i} \frac{H_j}{j+1}=\sum_{i=0}^{n-1}\frac{1}{i+1} \sum_{j=0}^{n-i-1} \frac{H_{j+1}}{j+2}=\sum_{i=0}^{n-1}\frac{1}{i+1} \sum_{j=i}^{n-1} \frac{H_{j+1-i}}{j+2-i}\\ =&\sum_{j=0}^{n-1}\sum_{i=0}^{j}\frac{H_{j+1-i}}{(i+1)(j+2-i)}=\sum_{j=0}^{n-1}\sum_{i=1}^{j+1}\frac{H_{j+2-i}}{i(j+3-i)}\\ =&\sum_{j=0}^{n-1}\sum_{i=1}^{j+1}\frac{1}{i(j+3-i)}\frac{1}{2}\left(\sum_{h=1}^{j+2-i}\frac{1}{h}+\sum_{h=1}^{j+2-i}\frac{1}{h}\right)\\ =&\sum_{j=0}^{n-1}\sum_{i=1}^{j+1}\frac{1}{i(j+3-i)}\frac{1}{2}\left(\sum_{h=1}^{j+2-i}\frac{1}{h}+\sum_{h=1}^{j+2-i}\frac{1}{j+3-i-h}\right)\\ =&\sum_{j=0}^{n-1}\sum_{i=1}^{j+1}\frac{1}{i(j+3-i)}\frac{1}{2}\sum_{h=1}^{j+2-i}\frac{j+3-i}{h(j+3-i-h)}=\frac{1}{2}\sum_{j=0}^{n-1}\sum_{i=1}^{j+1}\frac{1}{i}\sum_{h=1}^{j+2-i}\frac{1}{h(j+3-i-h)}\\ =&\frac{1}{2}\sum_{j=0}^{n-1}\sum_{i=1}^{j+1}\frac{1}{i}\sum_{h+g=j+3-i}\frac{1}{h\cdot g}=\frac{1}{2}\sum_{j=0}^{n-1}\sum_{h+g+i=j+3}\frac{1}{h\cdot g\cdot i}=\frac{1}{2}\sum_{j=1}^{n}\sum_{h+g+i=n-j+3}\frac{1}{h\cdot g\cdot i}\\ =&\frac{1}{2}\sum_{h+g+i+j=n+3}\frac{1}{h\cdot g\cdot i}=\frac{1}{2}\sum_{h+g+i+j=n+3}\frac{j}{h\cdot g\cdot i\cdot j} \end{align*}
But since the product $h\cdot g\cdot i\cdot j$ is unaffected by any permutation of the $4$-uple $(h,g,i,j)$ we have $\sum_{h+g+i+j=n+3}\frac{j}{h\cdot g\cdot i\cdot j}=\sum_{h+g+i+j=n+3}\frac{g}{h\cdot g\cdot i\cdot j}=\sum_{h+g+i+j=n+3}\frac{h}{h\cdot g\cdot i\cdot j}=\sum_{h+g+i+j=n+3}\frac{i}{h\cdot g\cdot i\cdot j}$ and then \begin{align*} \sum_{i=1}^{n}\frac{1}{n+1-i} \sum_{j=1}^i \frac{H_j}{j+1}=\frac{\frac{1}{2}\sum_{h+g+i+j=n+3}\frac{h+g+i+j}{h\cdot g\cdot i\cdot j}}{4}=\frac{n+3}{8}\sum_{h+g+i+j=n+3}\frac{1}{h\cdot g\cdot i\cdot j} \end{align*}
Comments: in the second line I do reindexing, in the third line I change the order of the sumations, in lines #4 #5 I replace the Harmonic number by its expression as a sum splitted in 2 parts, one of which is reindexed. Finally, line #7 is just reindexing.