Consider
$$\tag{1}f(x):=\sum_{n=1}^\infty \arctan\left(\left(\frac 14-n\right)x\right)-\arctan\left(\left(-\frac 14-n\right)x\right)$$
and let's rewrite the derivative of $f$ :
\begin{align}
\tag{2}f'(x)&=\frac 4{x^2}\sum_{n=1}^\infty\frac{1-4n}{(4n-1)^2+\bigl(\frac 4x\bigr)^2}-\frac{-1-4n}{(4n+1)^2+\bigl(\frac 4x\bigr)^2}\\
\tag{3}f'(x)&=\frac 4{x^2}\left(\frac {-1}{1^2+\left(\frac 4x\right)^2}+\sum_{k=1}^\infty\frac{k\sin\bigl(k\frac {\pi}2\bigr)}{k^2+\bigl(\frac 4x\bigr)^2}\right)\\
\end{align}
(since the $k=1$ term didn't appear in $(2)$)
But the series in $(3)$ may be obtained from $\,\frac d{d\theta} C_a(\theta)\,$ with :
$$\tag{4}C_a(\theta)=\frac {\pi}{2a}\frac{\cosh((\pi-|\theta|)a)}{\sinh(\pi a)}-\frac 1{2a^2}=\sum_{k=1}^\infty\frac{\cos(k\,\theta)}{k^2+a^2}$$
which may be obtained from the $\cos(zx)$ formula here (with substitutions $\ x\to\pi-\theta,\ z\to ia$).
The replacement of the series in $(3)$ by $\,\frac d{d\theta} C_a(\theta)\,$ applied at $\,\theta=\frac {\pi}2$ gives us :
\begin{align}
f'(x)&=\left(-\arctan\left(\frac x4\right)\right)'-\frac 4{x^2}C_{\frac 4x}\left(\theta\right)'_{\theta=\frac {\pi}2}\\
&=\left(-\arctan\left(\frac x4\right)\right)'+\frac{4\pi}{2x^2}\frac{\sinh\left(\frac{\pi}2 \frac 4x\right)}{\sinh\left(\pi\frac 4x\right)}\\
&=\left(-\arctan\left(\frac x4\right)\right)'+\frac{\pi}{x^2}\frac 1{\cosh\left(\frac{2\pi}x\right)}\\
\end{align}
Integrating both terms returns (with constant of integration $\frac {\pi}2$ since $f(0)=0$) :
$$f(x)=\frac {\pi}2-\arctan\left(\frac x4\right)-\arctan\left(\tanh\left(\frac {\pi}x\right)\right)\quad\text{for}\ \ x>0$$
i.e. the neat :
$$\tag{5}\boxed{\displaystyle f(x)=\arctan\left(\frac 4x\right)-\arctan\left(\tanh\left(\frac {\pi}x\right)\right)}\quad\text{for}\ \ x>0$$
So that your solution will be (for $x=1$) :
$$\boxed{\displaystyle \arctan(4)-\arctan\left(\tanh(\pi)\right)}$$
Your partial fraction decomposition is fine, but you need not to break up into separate positive and negative sums. Instead, you need to cancel terms within the summation.
$$\sum_{n=2}^{\infty}\bigg(\frac{1}{2(n+1)}+\frac{1}{2(n-1)}-\frac{1}{n}\bigg)=\\\frac 12\sum_{n=2}^{\infty}\bigg(\frac{1}{(n+1)}+\frac{1}{(n-1)}-\frac{2}{n}\bigg)=\\\frac 12\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}\right)-\left(\frac{1}{n}-\frac{1}{n+1}\right)$$
Now notice that the negative part of the $n$ term of the sum cancels with the positive part of the $n+1$ term of the sum, so all the terms disappear except the positive part of the $n=2$ term. Our sum then equals
$$\frac 12\left(\frac 11-\frac 12\right)=\frac 14$$
Best Answer
There are more terms surviving after telescoping. Namely, we are left with \begin{align}\lim_{n \rightarrow \infty} \Bigl(\arctan (n+1) + \arctan n - \arctan 1 - \arctan 0\Bigr) = \frac{ \pi}{2} +\frac{ \pi}{2} -\frac{ \pi}{4}-0 =\frac{ 3\pi}{4}. \end{align}