Consider the $4 \times 4$ grid. Place $4$ overlapping copies of $G$ as shown (left figure), where the number $i = 1, 2, 3, 4$ indicates the cells of the $i^{\text{th}}$ copy of $G$. The same grid is also covered by $8$ overlapping copies of $2 \times 2$ grid (right), with each cell covered the same number of times. (Note that in the first figure the top left cells of the $4$ copies of $G$ form a $2 \times 2$ grid while in the second figure, the top left cells of the $8 \times 2 \times 2$ grid form the configuration $G$. This is important for the general case.)

Thus the sum of the sums of the numbers in the cells of the $4$ copies of $G$ is equal to that of the $8$ copies of the $2 \times 2$ grid. Since the former is positive, one of the $2 \times 2$ grid must be positive as well.

It is easy to see that the result holds for any two configurations $G_1, G_2$, provided the grid is large enough. Choose a pair of coordinate axes. Let the number of cells in $G_1$ and $G_2$ be $m$ and $n$ respectively. Place $G_1$ and $G_2$ in some position. Let $a_1, \dots, a_m$ be the centres of the cells covered by $G_1$ and let $b_1, \dots, b_n$ be the centres of the cells covered by $G_2$. Let $G_1(b_i)$ be the position of $G_1$ after it has been translated by $b_i$ and let $P_1(b_i)$ be the corresponding sum. The sum $P_2(a_i)$ is similarly defined. It is clear that the cells, counting multiplicity, covered by $G_1$ when it is translated by the vectors $b_1, b_2, \dots, b_n$ are the same as the cells, counting multiplicity, covered by $G_2$ when it is translated by the vectors $a_1, a_2, \dots, a_m$. Thus
$$\sum P_1(b_i) = \sum P_2(a_j).$$
Since the LHS is positive, at least one of the terms on the RHS is positive.