From the following problem, I think it is not possible:
" You find two signatures made by Alice. You know that she is using the ElGamal signature scheme over $\mathbb{F}_{2027}$. The cyclic group $\mathbb{G}$ she is using is a (multiplicative) subgroup of order 1013. The signatures are on hash values $h(m_1) = 345$ and $h(m_2) = 567$ and are given by $(r_1, s_1) = (365, 448)$ and $(r_2, s_2) = (365, 969)$. Compute (a candidate for) Alice’s secret key $x$ based on these signatures, i.e. break the system. "
By the following equations, this is why I think these combinations are not possible:
$\begin{cases} (365, 448) = (g^k \mod 2027 , (345-365x)k^{-1} \mod 2026) \\ (365, 969) = (g^k \mod 2027, (567-365x)k^{-1} \mod 2026) \end{cases}$
The difference of the $y$-coordinate is $969-448 = 521\mod 2026$ from the left side, and $(567-345)k^{-1}=222k^{-1} \mod 2026$ from the right side. So the left side is odd and the right side is even while there are equal.
Do I do something wrong, or do you agree?
The verifying equations are
$$H(m) = xr + ks \bmod{(p-1)}\tag{1}$$
and
$$H(m') = xr + ks' \bmod{(p-1)}\tag{2}$$
and as we have a common $k$ and $r$, there are two unknowns $x$ and $k$.
Solve these equations by standard techniques, $H(m), H(m')$ are known and find $x$ (and $k$ too, but that's not very useful).
I did and found an $x$ that works. If $k$ works so does $k+1012$ as $g$ has order $1012$, and so these give the same $r$.