The total resistance of the grid is infinite when the grid is two dimensional and large.
If you place two point probes at location x and y on an infinite 2-d resistor grid, and impose the voltage V(x)=1 and V(y)=0, the potential obeys the discretized Laplace equation: V(up) + V(down) + V(left) + V(right) - 4 V(center) = 0 with the boundary conditions at the two given points and V=0 at infinity (beyond x and y).
In the limit that x and y are far apart, the discrete Laplace equation might as well be the continuous Laplace equation, and the solution goes like C log(|r-x|/|y-x|), so that the potential difference for any finite C diverges with the distance. This means that C has to go to zero in the large |x-y| limit, so the current vanishes. The same is true in 1d, where a line of resistors has a current which vanishes as 1/L, so the total resistance goes as the total length L. In two dimensions, the total resistance blows up as log(L).
For a three dimensional grid and higher, you do have a finite resistance for a block. Whether the limiting resistance is finite or infinite is the same problem as the recurrence/nonrecurrence of a random walk on the grid.
If you make a pseudo two-d grid using N parallel lines of N resistors in series, then the total resistance is N on each path, but there are N parallel paths, so the total resistance is R, independent of the size. This is not the same as the 2-d resistor grid, because in the 2d grid there is resistance to going vertically a long way which is similar to the resistance to going horizontally, so the horizontal resistor paths are not parallel. If you make all the vertical resistors zero, and make the separation between x and y horizontal, and make the vertical width equal to |x-y|, you recover the series/parallel situation.
The series-parallel example gives intuition about why two dimensions is critical for the transition from infinite resistance to finite resistance.
While mmc's answer is correct, the result of applying Kirchhoff's laws is that you get the Graph Laplacian problem.
Given a graph $G$ with each edge $E_i$ given a resistance $R_i$, the weighted graph Laplacian is given by considering the operation on functions which takes $\phi(V)$ to
$$ \nabla^2 \phi = \sum_{<W,V>} {1\over R_i} (\phi(W)-\phi(V)) $$
Where the sum extends over all $W$ which are graph neighbors of $V$. This graph Laplacian then can be solved with discrete delta-function (kronecker delta) sources. The interpretation of the Laplacian is that $\phi$ is the voltage at each point, the Laplacian is summing the currents entering each node (with sign) and setting it to zero.
If you force a certain current into one node and extract the same current at another node, you are solving the equation
$$ \nabla^2 \phi = \delta_{V,V_1} - \delta_{V,V_2} $$
The solution gives the electrostatic potential configuration for all the nerd-sniping problems. For finite graphs, this is a matrix inversion. For infinite homogenous graphs with a translation symmetry, you can reduce the translation invariant directions by Fourier transforms, which allows you to solve any homognenous grid. For general graphs, you study the graph Laplacian, which is related to the random walk on the same graph, which has many bounds on the eigenvalues, which give you the asymptotic behavior of the potential in nearly all the cases of interest.
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Best Answer
Yes, it is possible. For example Kevin Brown did here and here including this table.
so for the xkcd problem the answer is $-\frac{1}{2}+\frac{4}{\pi} \approx 0.773$.