An interesting sum over lattice points in a large disk centered at the origin

The limit equals $-\pi\log 2$, in accordance with Henri Cohen's remark above.

For the proof, we combine the formula $$r_2(k):=\#\{(m,n)\in\mathbb{Z}^2:m^2+n^2=k\}=4\sum_{d\mid k}\chi_4(d)$$ with Dirichlet's hyperbola method. With this notation, the OP's sum equals $$\sum_{k\leq r^2}\frac{(-1)^kr_2(k)}{k}=4\sum_{de\leq r^2}\frac{(-1)^{e}\chi_4(d)}{de}.$$ We shall now use the well-known facts that $$\sum_{d=1}^\infty\frac{\chi_4(d)}{d}=\frac{\pi}{4} \qquad\text{and}\qquad \sum_{e=1}^\infty\frac{(-1)^e}{e}=-\log 2,$$ where both series are alternating. Using this observation, \begin{align*} \sum_{de\leq r^2}\frac{(-1)^{e}\chi_4(d)}{de} =&\sum_{\substack{d\leq r\\e\leq r^2/d}}\frac{(-1)^{e}\chi_4(d)}{de} +\sum_{\substack{e\leq r\\d\leq r^2/e}}\frac{(-1)^{e}\chi_4(d)}{de}- \sum_{d,e\leq r}\frac{(-1)^{e}\chi_4(d)}{de}\\ =&\sum_{d\leq r}\frac{\chi_4(d)}{d}\left(-\log 2+O\left(\frac{d}{r^2}\right)\right)\\ &+\sum_{e\leq r}\frac{(-1)^e}{e}\left(\frac{\pi}{4}+O\left(\frac{e}{r^2}\right)\right)\\ &-\left(\sum_{d\leq r}\frac{\chi_4(d)}{d}\right) \left(\sum_{e\leq r}\frac{(-1)^e}{e}\right)\\ =&-\log 2\sum_{d\leq r}\frac{\chi_4(d)}{d}+\frac{\pi}{4}\sum_{e\leq r}\frac{(-1)^e}{e}+\frac{\pi}{4}\log 2+O\left(r^{-1}\right)\\ =&-\frac{\pi}{4}\log 2-\frac{\pi}{4}\log 2+\frac{\pi}{4}\log 2+O\left(r^{-1}\right)\\ =&-\frac{\pi}{4}\log 2+O\left(r^{-1}\right). \end{align*} Therefore, the OP's sum is $$\sum_{k\leq r^2}\frac{(-1)^kr_2(k)}{k}=-\pi\log 2+O\left(r^{-1}\right).$$

It is problem number 10 of IMC 2018, you may find the solution on the official site.

More generally, when $\Re(s)\ge1$ $$\sum_{(m,n)\ne(0,0)}(-1)^{m+n}/(m^2+n^2)^s=-4(1-2^{1-s})\zeta_K(s)$$ where $K=\mathbb Q(i)$ and $\zeta_K$ is the Dedekind zeta function of $K$.