The entanglement negativity $\mathcal N(\rho)$ of a (bipartite) state $\rho$ is defined as the absolute value of the sum of the negative eigenvalues of the partial transpose of a state, or equivalently, $2\mathcal N(\rho)=\|\rho^{T_B}\|_1-1$.
Consider a maximally entangled pure state: $|\psi\rangle \simeq \sum_{k=1}^N |k,k\rangle$. What is the entanglement negativity of such states? What are good ways to derive this quantity?
Let $\sqrt N|\psi\rangle=\sum_{k=1}^N|k,k\rangle$ be a maximally entangled state of dimension $N$, and $\rho\equiv|\psi\rangle\!\langle\psi|$. More generally, we don't need to stick to maximally entangled states: any state with this (or equivalent) Schmidt decomposition will behave identically.
The partial transpose reads $\rho^{T_B}=\frac{1}{N}\sum_{ij}\lvert ij\rangle\!\langle ji\rvert. $ Separating this into terms that are symmetric and terms that are not we get $$N\rho^{T_B} = \sum_i \lvert ii\rangle\!\langle ii| + \underbrace{\sum_{i\neq j}\lvert ij\rangle\!\langle ji\rvert}_{A},$$ where here $A^\dagger =A$, $\operatorname{tr}(A)=0$, and $A^2=I$. It follows that $A$ has an equal number of eigenvalues equal to $+1$ and $-1$. Moreover, the rank of $A$ is $N(N-1)$, so the multiplicity of both eigenvalues is $N(N-1)/2$.
The eigenvalues of $\rho^{T_B}$ are therefore $\frac1 N$ with multiplicity $N+\frac{N(N-1)}2=\frac{N(N+1)}{2}$, and $-\frac1 N$ with multiplicity $\frac{N(N-1)}2$.
We conclude that the negativity equals $\frac{N(N-1)}{2}\frac{1}{N}=\frac{N-1}{2}$.