For which $c\in\mathbb{R}$ there exists $t\in\mathbb{R}^+$ such that $x_1(t)=0$ and $x_2(t)>0$ for the system $x_1'(t)=-200kx_2(t)-c\ ,\ x_2'(t)=-100kx_1(t)\ ,\ x_1(0)=10^4\ ,\ x_2(0)=5\cdot 10^3\ ,\ k>0$.
What I did: I found this form of $X(t)$: $$\begin{pmatrix}x_1(t)\\x_2(t)\end{pmatrix} =(\frac{c}{400\sqrt2k}+2500(1+\sqrt2))e^{-100\sqrt2kt}\begin{pmatrix}\sqrt2\\1\end{pmatrix}+(\frac{c}{400\sqrt2k}+2500(1-\sqrt2))e^{100\sqrt2kt}\begin{pmatrix}-\sqrt2\\1\end{pmatrix} -\begin{pmatrix}0\\\frac{c}{200k}\end{pmatrix}$$
If $x_1=0$ we get: $$t=\frac{10^2\cdot ln(\frac{c+10^6\cdot\sqrt2k(1+\sqrt2)}{c+10^6\cdot\sqrt2k(1-\sqrt2)})}{2\sqrt2k}$$
Which means that for $x_1$ to be able to reach $0$ we must have: $$c>10^6\cdot\sqrt2k(\sqrt2-1)$$
I wasn't able to simplify the expression for $x_2=0$ and arrive at a final answer.
You already expressed $t$ (such that $x_1(t)=0$ in terms of $c$). By the way, there should be
$$t=\frac{ln\left(\frac{c+10^6\cdot\sqrt2k(1+\sqrt2)}{c+10^6\cdot\sqrt2k(1-\sqrt2)}\right)}{200\sqrt2k}.$$
So it remains to make the respective substitution in the formula for $x_2(t)$ and then to find when $x_2(t(c))>0$.
That is
$$\left(\frac{c}{400\sqrt2k}+2500(1+\sqrt2)\right)\sqrt{\frac{c+10^6\cdot\sqrt2k(1-\sqrt2)}{c+10^6\cdot\sqrt2k(1+\sqrt2)}}+$$ $$\left(\frac{c}{400\sqrt2k}+2500(1-\sqrt2)\right)\sqrt{\frac{c+10^6\cdot\sqrt2k(1+\sqrt2)}{c+10^6\cdot\sqrt2k(1-\sqrt2)}}>\frac{c}{200k}$$
$$\left(c+10^6\sqrt2k(1+\sqrt2)\right)\sqrt{\frac{c+10^6\cdot\sqrt2k(1-\sqrt2)}{c+10^6\cdot\sqrt2k(1+\sqrt2)}}+$$ $$\left(c+10^6\sqrt2k(1-\sqrt2)\right)\sqrt{\frac{c+10^6\cdot\sqrt2k(1+\sqrt2)}{c+10^6\cdot\sqrt2k(1-\sqrt2)}}>2\sqrt2c$$
$$2\sqrt{(c+10^6\cdot\sqrt2k(1-\sqrt2))(c+10^6\cdot\sqrt2k(1+\sqrt2)})>2\sqrt2c$$
$$(c+10^6\cdot\sqrt2k(1-\sqrt2))(c+10^6\cdot\sqrt2k(1+\sqrt2))>2c^2$$
$$10^6\cdot2\sqrt2kc-10^{12}\cdot 2k^2>c^2$$
$$c^2-10^6\cdot2\sqrt2kc+10^{12}\cdot 2k^2<0$$
$$(c-10^6\sqrt2k)^2<0,$$
which is impossible.
PS. This simple answer suggests that there may be a more simple solution.