Proof for values of d with d:= ||L|| - N(L) with $d \in \mathbb{Z}$ and N(L) Nerode Index Let ||L|| be the sum of all lengths of words in L und N(L) the number of equivalence claesses for the Relation $\equiv_L$ from Myhill–Nerode theorem. Proof, which values d can have with $d:=||L||-N(L),d\in\mathbb{Z}$
Some conclusions I came up with so far:

|L|| is infinite, when L is infinite. Then d is not defined.
When N(L) is infinite for any non regular language, so L is infinite as well and $\infty - \infty $ wouldn´t be defined.
So say ||L|| is finite then N(L) is finite as well because every word hase to be in a certain class. If not ||L|| would be infinite. Though when N(L) is finite, ||L|| doesn´t have to be finite since we might have some symbols looping over states.
d is probably never negative, because if N(L) &gt; ||L|| we would have classes which don´t belong to words in L.
When d = 0 then, N(L) = ||L|| and there would be a class for every word in L.

So d can be anything from $\{ 0,...,\infty \}$
Might this be enough for an answer or am I missing any cases/details? Where could I be proofing more formally?
In order to make the question nontrivial, you have to assume that $L$ is finite. You haven't specified what the alphabet is, so I will assume that the alphabet is arbitrary.
First, let us show that $N(L) \leq \|L\|+2$. We can construct a DFA for $L$ whose states consist of all prefixes of words in $L$, together with a failure state. This DFA contains an initial state (corresponding to the empty prefix), a failure state, and one state for every non-empty prefix. In particular,
$$
N(L) \leq 2 + \sum_{w \in L} |w| = 2 + \|L\|.
$$
In other words, $d \geq -2$.
Second, let us consider the language $L_\Sigma = \{\sigma : \sigma \in \Sigma\}$ over an alphabet $\Sigma$. The minimal DFA for this language contains three states, and so the value of $d$ for $L_\Sigma$ is
$$
\|L_\Sigma\| - N(L_\Sigma) = |\Sigma| - 3.
$$
As $\Sigma$ ranges over all possible alphabets, we obtain all values which are at least $-2$.