\begin{align} S(r,s) = \sum_{k=0}^{\infty} = \frac{(-1)^k}{(2k+r+1)(2k+s+1)} \end{align} Assuming $r>s\geq 0$ and both $r,s$ even integers.
A few values, using wolfram:
\begin{align} &S(2,0) = \pi/4-1/2\\ &S(4,0) = 1/6\\ &S(4,2) = 5/6 - \pi/4\\ &S(6,0) = \pi/12-13/90\\ &S(6,2) = 1/30\\ &S(6,4) = \pi/4 - 23/30\\ &S(8,0) = 19/210\\ &S(8,2) = 181/630 - \pi/12 \end{align}
For the rational values of $S(r,s)$ we have:
\begin{align} S(r,s) = \frac{1}{r-s} \left( \frac{1}{s+1}-\frac{1}{s+2}+\dots+\frac{1}{r-1} \tag{1} \right) \end{align}
I think for the irrational values must be something similar to $(1)$ except we will have to add $\pi$ after the sum.
What is the general formula for $S$ (or just the formula for irrational values) $?$
Define $$S_n(r,s):=\sum_{k=0}^n\,\frac{(-1)^k}{(2k+r+1)(2k+s+1)}$$ for each $n=0,1,2,\ldots$. Write $r=2p$ and $s=2q$ for some integers $p$ and $q$ such that $p>q\geq 0$. Thus $$S_n(2p,2q)=\sum_{k=0}^n\,\frac{(-1)^k}{2(p-q)}\,\left(\frac{1}{2k+2q+1}-\frac{1}{2k+2p+1}\right)\,.$$ That is, $$S_n(2p,2q)=\frac{1}{2(p-q)}\,\left(\sum_{k=0}^n\,\frac{(-1)^k}{2k+2q+1}-\sum_{k=0}^n\,\frac{(-1)^k}{2k+2p+1}\right)\,.$$ Note that, for each nonnegative integer $t$, $$\sum_{k=0}^n\,\frac{(-1)^k}{2k+2t+1}=(-1)^{t-1}\,\sum_{k=0}^{t-1}\,\frac{(-1)^k}{2k+1}+(-1)^t\,\sum_{k=0}^{n+t}\,\frac{(-1)^k}{2k+1}\,.$$ Hence, $$S_n(2p,2q)=\small\frac{1}{2(p-q)}\,\left((-1)^{q-1}\,\sum_{k=0}^{q-1}\,\frac{(-1)^k}{2k+1}-(-1)^{p-1}\,\sum_{k=0}^{p-1}\,\frac{(-1)^k}{2k+1}\right)+\frac{(-1)^q\,T_{n+q}-(-1)^p\,T_{n+p}}{2(p-q)}\,,$$ where $T_m:=\sum\limits_{k=0}^m\,\dfrac{(-1)^k}{2k+1}$ for $m=0,1,2,\ldots$. Note that $\lim\limits_{m\to\infty}\,T_m=\dfrac{\pi}{4}$. Ergo, $$\begin{align}S(2p,2q)&=\lim_{n\to\infty}\,S_n(2p,2q)\\&=\small\frac{1}{2(p-q)}\,\left((-1)^{q-1}\,\sum_{k=0}^{q-1}\,\frac{(-1)^k}{2k+1}-(-1)^{p-1}\,\sum_{k=0}^{p-1}\,\frac{(-1)^k}{2k+1}\right)+\left(\frac{(-1)^q-(-1)^p}{2(p-q)}\right)\,\frac{\pi}{4}\,.\end{align}$$
That is, $S(2p,2q)$ is rational if and only if $p\equiv q\pmod{2}$. In this case, we have $$S(2p,2q)=\frac{(-1)^{q}}{2(p-q)}\,\sum_{k=q}^{p-1}\,\frac{(-1)^k}{2k+1}=\small\frac{1}{2(p-q)}\,\left(\frac{1}{2q+1}-\frac{1}{2q+3}+\frac{1}{2q+5}-\ldots-\frac{1}{2p-1}\right)\,.$$ On the other hand, if $p\not\equiv q\pmod{2}$, we obtain $$S(2p,2q)=\frac{(-1)^q}{2(p-q)}\,\left(\frac{\pi}{2}-\left(\sum_{k=0}^{q-1}\,\frac{(-1)^k}{2k+1}+\sum_{k=0}^{p-1}\,\frac{(-1)^k}{2k+1}\right)^{\vphantom{a^a}}_{\vphantom{a_a}}\right)\,.$$ However, note also that $$S(2q,2q)=(-1)^q\,\left(C-\,\sum_{k=0}^{q-1}\,\frac{1}{(2k+1)^2}\right)\,,$$ where $C:=\sum\limits_{k=0}^\infty\,\frac{(-1)^k}{(2k+1)^2}$ is Catalan's constant.
For integers $p\geq q\geq 0$, we also have $$S_n(2p+1,2q)=\sum_{k=0}^n\,\frac{(-1)^k}{2p-2q+1}\,\left(\frac{1}{2k+2q+1}-\frac{1}{2(k+p+1)}\right)\,.$$ Therefore, $$\begin{align}S_n(2p+1,2q)&=\small\frac{1}{2p-2q+1}\,\left((-1)^q\,T_{n+q}+(-1)^{q-1}\,\sum_{k=0}^{q-1}\,\frac{(-1)^k}{2k+1}+\frac{(-1)^{p-1}}{2}\,R_{n+p}+\frac{(-1)^p}{2}\,\sum_{k=0}^{p-1}\,\frac{(-1)^k}{k+1}\right)\,,\end{align}$$ where $$R_m:=\sum_{k=0}^m\,\frac{(-1)^k}{k+1}\,.$$ Because $\lim\limits_{m\to\infty}\,R_m=\ln(2)$, we get that $$\begin{align}S(2p+1,2q)&=\small\frac{(-1)^q}{2(2p-2q+1)}\,\left(\frac{\pi}{2}-(-1)^{p-q}\,\ln(2)-2\,\sum_{k=0}^{q-1}\,\frac{(-1)^k}{2k+1}+(-1)^{p-q}\,\sum_{k=0}^{p-1}\,\frac{(-1)^k}{k+1}\right)\,.\end{align}$$
Likewise, for integers $p$ and $q$ with $p>q\geq 0$, we have $$S_n(2p,2q+1)=\sum_{k=0}^n\,\frac{(-1)^k}{2p-2q-1}\,\left(\frac{1}{2(k+q+1)}-\frac{1}{2k+2p+1}\right)\,.$$ Ergo, $$\begin{align}S_n(2p,2q+1)&=\small\frac{1}{2p-2q-1}\,\left(\frac{(-1)^q}{2}\,R_{n+q}+\frac{(-1)^{q-1}}{2}\,\sum_{k=0}^{q-1}\,\frac{(-1)^k}{k+1}+(-1)^{p-1}\,T_{n+p}+(-1)^p\,\sum_{k=0}^{p-1}\,\frac{(-1)^k}{2k+1}\right)\,,\end{align}$$ whence $$S(2p,2q+1)=\small\frac{(-1)^q}{2(2p-2q-1)}\,\left(\ln(2)-(-1)^{p-q}\,\frac{\pi}{2}-\sum_{k=0}^{q-1}\,\frac{(-1)^k}{k+1}+2(-1)^{p-q}\,\sum_{k=0}^{p-1}\,\frac{(-1)^k}{2k+1}\right)\,.$$
For integers $p>q\geq 0$, we have $$S_n(2p+1,2q+1)=\sum_{k=0}^n\,\frac{(-1)^k}{4(p-q)}\,\left(\frac{1}{k+q+1}-\frac{1}{k+p+1}\right)\,.$$ That is, $$S_n(2p+1,2q+1)=\frac{1}{4(p-q)}\,\left(\sum_{k=0}^n\,\frac{(-1)^k}{k+q+1}-\sum_{k=0}^n\,\frac{(-1)^k}{k+p+1}\right)\,.$$ Hence, $$\begin{align}S_n(2p+1,2q+1)&=\small\frac{1}{4(p-q)}\,\left((-1)^{q-1}\,\sum_{k=0}^{q-1}\,\frac{(-1)^k}{k+1}-(-1)^{p-1}\,\sum_{k=0}^{p-1}\,\frac{(-1)^k}{k+1}\right)+\frac{(-1)^q\,R_{n+q}-(-1)^p\,R_{n+p}}{4(p-q)}\,,\end{align}$$ Ergo, $$\begin{align}S(2p+1,2q+1)&=\small\frac{1}{4(p-q)}\,\left((-1)^{q-1}\,\sum_{k=0}^{q-1}\,\frac{(-1)^k}{k+1}-(-1)^{p-1}\,\sum_{k=0}^{p-1}\,\frac{(-1)^k}{k+1}\right)+\left(\frac{(-1)^q-(-1)^p}{4(p-q)}\right)\,\ln(2)\,.\end{align}$$
That is, $S(2p+1,2q+1)$ is rational if and only if $p\equiv q\pmod{2}$. In this case, we have $$\begin{align}S(2p+1,2q+1)&=\frac{(-1)^{q}}{4(p-q)}\,\sum_{k=q}^{p-1}\,\frac{(-1)^k}{k+1}=\small\frac{1}{4(p-q)}\,\left(\frac{1}{q+1}-\frac{1}{q+2}+\frac{1}{q+3}-\ldots-\frac{1}{p}\right)\,.\end{align}$$ On the other hand, if $p\not\equiv q\pmod{2}$, we obtain $$\begin{align}S(2p+1,2q+1)&=\frac{(-1)^q}{4(p-q)}\,\left(2\,\ln(2)-\left(\sum_{k=0}^{q-1}\,\frac{(-1)^k}{k+1}+\sum_{k=0}^{p-1}\,\frac{(-1)^k}{k+1}\right)^{\vphantom{a^a}}_{\vphantom{a_a}}\right)\,.\end{align}$$ However, note also that $$S(2q+1,2q+1)=\frac{(-1)^q}{4}\,\left(\frac{\pi^2}{12}-\,\sum_{k=0}^{q-1}\,\frac{1}{(k+1)^2}\right)\,.$$