Current through each resistor
Extending Olin's answer for those that want to see how much of a difference the art makes.







simulate this circuit &ndash; Schematic created using CircuitLab

Once you see it that way it's pretty easy to tell, or should be, what voltage those nodes in the middle are at. After that it's simple Ohm's Law stuff.

You should also be able to see that the nodes I marked as \$VT\$ have the same voltage, as do those I marked with \$VB\$. 

Since there is no voltage across \$R6\$ and \$R9\$, 

\$I6 = I9 = 0A\$.

Because of that, \$R6\$ and \$R9\$ can be considered as shorts, so \$R4\$ is effectively in parallel with \$R2\$ (\$R24 = 50R\$). Which means \$VT\$ is \$50/150\$ or \$1/3\$ of \$12.5V\$ above the mid-point, that is, 

\$VT = 12.5 + 12.5/3 = 16.667V\$

Similarly \$R8\$ is effectively in parallel with \$R10\$, making 

\$VB = 2/3 * 12.5V = 8.333V\$.

So, 

\$I1 = I3 = I12 = I14 = 8.333/100 = 83.33mA\$
\$I2 = I4 = I5 = I7 = I8 = I13 = I11 = I10 = I1/2 = 41.667mA\$

\$Io = I1 + I2 = 166.67mA\$

The puzzle is actually delightful in it's symmetry. Of course this only falls out like this because all the resistor values are identical. Had they not been, your original method, or some other analysis method, would be required.