Actually, the contours $C$ in the two integrals are different.


By definition, $C$ goes from $+\infty$ through an anticlockwise path return to $+\infty$, encircling all poles of $\Gamma(b_j-s)$ (exactly once) and none pole of $\Gamma(1-a_j-s)$. In this case, the integral-contours of $\textrm{G}^{4, 0}_{2, 4}\Big({1,c+\frac12; c,c,c,1}\Big|1\Big)$ encircles poles:

$$c, c+1, c+2,\dotsc, c+n,\dotsc,$$

and poles:

$$0, -1, -2, \dotsc, -n, \dotsc,$$

must be outside the contours. Similar for $\textrm{G}^{3, 1}_{2, 4}\Big({1,c+\frac12; c,c,c,1}\Big|1\Big)$, we needs the contour encircles poles:

$$0, 1, 2, \dotsc, n,\dotsc,\quad \mathrm{and}\quad c, c+1,c+2,\dotsc,c+n,\dotsc,$$

As you have found, $1,2,\dotsc,n,\dotsc$ are actually normal points of the integrand, so the contour can "deform" over these points. However, $0$ remains to be a pole, so the contour can not cross over it, which makes the expression obtain a non-zero value.

Thus, for any $c\neq0,-1,-2,\dotsc,-n,\dotsc$ (by definition, they're not allowed), the sum of this two Meijer functions is:

$$-\mathrm{Res}\left(\frac{\Gamma(c-s)^3}{s\Gamma(c+\frac12-s)}\right)_{s=0}
    =-\dfrac{\Gamma(c)^3}{\Gamma(\frac12+c)}.$$

Using Mathematica, we can check our answer:

```
Table[-MeijerG[{{},{1,c+1/2}},{{0,c,c,c},{}},1,-1]
	-MeijerG[{{1},{c+1/2}},{{c,c,c},{0}},1,-1]
	+Gamma[c]^3/Gamma[1/2+c]//N, 
	{c, SetPrecision[RandomReal[{1,2}, 10]
        +I RandomReal[{1, 2}, 10], 10]}
]//Column
```

[![Results][1]][1]
Notice that in Mathematica
```
-MeijerG[{{__}, {__}}, {{__}, {__}}, _, -1]
```

is corresponding to the definition in Wikipedia(Bateman & Erdelyi).

  [1]: https://i.sstatic.net/t8U1W.png