Prove that the lower bound for the partial sums of the Möbius inverse of the Harmonic numbers minus $n$ is greater than $1$ minus a Harmonic number Let $\varphi^{-1}(n)$ be the Dirichlet inverse of the Euler totient function:
$$\varphi^{-1}(n) = \sum\limits_{d\mid n} d \cdot \mu(d) \tag{1}$$
and let the matrix $T$ be:
$$T(n,k)=\varphi^{-1}(\gcd(n,k)) \tag{2}$$
Because:
$$\Lambda(n) = \sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k}, \tag{3}$$
therefore the Möbius inverse of the Harmonic numbers $H_{n}^{-1}$ is:
$$H_{n}^{-1}=\sum _{k=1}^{n} \frac{\varphi^{-1}(\gcd (n,k))}{k} \tag{4}$$
The partial column sums of the lower triangular part of $T(n,k)$ is:
$$M(n,k)=\underset{m\geq k}{\sum _{m=1}^n} \varphi^{-1}(\gcd (m,k)) \tag{5}$$
The partial sums of the Möbius inverse of the Harmonic minus $n$ is therefore:
$$-n +\sum_{n=1}^{N} H_{n}^{-1} = \sum_{k=2}^{n} \frac{M(n,k)}{k} \tag{6}$$

Conjecture and question:
  Show that:
$$|M(p_i-1,k)|>0 \tag{7}$$
       where $p_i$ is the $i$-th prime number, $i=1,2,3,4,5,...$.
   for all $1 \leq k \leq p_i-1$.   

Consider the linear programming problems:
$$\begin{array}{ll} \text{minimize} & \displaystyle\sum_{k=1}^{n} \frac{x_{k}}{k} \\ \text{subject to constraints:} & n + \displaystyle\sum_{k=2}^{n}x_{k}=1 \\ & x_1 \geq 1 \\ k>1: & -1 \leq x_k \leq 1 & \tag{8} \end{array}$$
where the objective function above to be minimized is what we are seeking to (lower) bound in RHS of $(6)$ for $k=1,2,3,4,5,...,n$:
$$\sum_{k=1}^{n} \frac{x_{k}}{k} \iff \sum_{k=2}^{n} \frac{M(n,k)}{k} \tag{9}$$
and where the constraint is the property:
$$n + \sum_{k=1}^{n} \frac{x_{k}}{k} = 1 \iff \sum_{k=1}^{k=n} M(n,k) = 1 \tag{10}$$
proven here by Max Alekseyev.
and where the variable bounds are:
$$ k=1: \; x_1 \geq 1 \iff M(n,1)=n  \tag{11}$$
$$ k>1: \; -1 \leq x_k \leq 1 \iff |M(p_i-1,k)|>0 \tag{12}$$
The solutions to linear programming problem $(8)$ is:
$$1+\sum _{k=1}^n -\frac{1}{k} \tag{13}$$
Which depends on the conjecture in $(7)$ and which means that lower bound of RHS of $(6)$ is:
$$\text{lower bound}\left(\sum_{k=2}^{n} \frac{M(n,k)}{k}\right)<1+\sum _{k=1}^n -\frac{1}{k} \tag{14}$$
which in turn is:
$$\text{lower bound}\left(-n +\sum_{n=1}^{N} H_{n}^{-1}\right)<1+\sum _{k=1}^N -\frac{1}{k} \tag{15}$$
which is the question in the title:
"Prove that the lower bound for the partial sums of  the Möbius inverse of the Harmonic numbers minus $n$ must be more than $1$ minus a Harmonic number, when $n=p-1$ for $p$ a prime number."
 A: This answer is not complete yet.
It is known that the $k$-th columns in:
$$\gcd(1..k+1,k) \tag{1}$$
and:
$$\gcd(k..2k+1,k) \tag{2}$$
are palindromic for all $k$ and therefore the $k$-th columns in:
$$T(1..k+1,k) \tag{3}$$
and:
$$T(k..2k+1,k) \tag{4}$$
are also palindromic for all $k$.
$$T(n,k)=\varphi^{-1}(\gcd(n,k)) \tag{5}$$ is as stated in the question.
Since the sequences in the columns are periodic and periods sum to zero, the palindromicity means that for $n=1..k$ the $k$-th columns of the sum:
$$\sum_{m=1}^{n} \varphi^{-1}(\gcd(m,k)) \tag{6}$$
for $n=(k-1)..2(k-1)$ are the reversed sequences of the $k$-th columns of the negated sum:
$$-\sum_{m=1}^{n} \varphi^{-1}(\gcd(m,k)) \tag{7}$$
for all $k>1$.
Mathematica program to demonstrate that palindromicity gives negated reversals of columns.
This is as far as I got. The next step must have something to with the cases when:
$$T(n,k)=\varphi^{-1}(\gcd(n,k))=1 \tag{8}$$
and how such cases can't cause a zero in:
$$\sum_{m=1}^{n} \varphi^{-1}(\gcd(m,k))$$ when $n=1,2,3,4,5,...$ is a prime number, and how the reversal is shifted by one step.
But I don't know how to prove it.

Edit 29.3.2020:
Let 
$$T(n,k)=\varphi^{-1}(\gcd(n,k))  \tag{9}$$
The following holds:
$$\varphi^{-1}(1)=1  \tag{10}$$
$$T(1,k)=\varphi^{-1}(\gcd(1,k))=1  \tag{11}$$
for all $k \geq 2$.
$$\sum\limits_{m=1}^{k} T(m,k) = \sum\limits_{m=1}^{k} \varphi^{-1}(\gcd(m,k)) = 0  \tag{12}$$
for all $k \geq 2$.
Let: 
$$W(n,k)=\sum\limits_{m=1}^{n} \varphi^{-1}(\gcd(m,k))  \tag{13}$$
Since prime numbers $p_i$ are not multiples of any $k$ and 
$$\varphi^{-1}(n) \neq 0 \tag{14}$$
for any $n$ and in particular $n=1$,
and since it is possible for some $k$ that:
$$\sum\limits_{m=1}^{m_1} \varphi^{-1}(\gcd(m,k))=0  \tag{15}$$
only for some $1<m_1 \leq k$,  
and since the columns in $W(n,k)$ are periodic for $k>1$:
$$\sum\limits_{m=1}^{n} \varphi^{-1}(\gcd(m,k))=\sum\limits_{m=1}^{n+k} \varphi^{-1}(\gcd(m,k))  \tag{16}$$
which is the same as:
$$W(n,k)=W(n+k,k) \tag{17}$$
Therefore
$$\sum\limits_{m=1}^{p_i} \varphi^{-1}(\gcd(m,k)) \neq 0  \tag{18}$$
for $$k=1..(p_i-1) \tag{19}$$
I am unsatisfied with having $k$ both in the upper summation limit in (12) and in the argument in (19), because it is confusing. This is the best I can do at the moment.
