The key principle of this problem is symmetry.
[The other key principle I have demonstrated is not being lazy about a problem that looks easy! I previously got sloppy and tried to make it TOO simple. I believe I've corrected the errors.]
I would further say that the purpose of this problem is precisely to make you think clearly about the circuit, rather than try to solve it mathematically. I'm sad to hear that you didn't learn techniques for drawing clear circuit diagrams in school, because that's maybe the most useful practical skill in solving this problem. (When I saw your question, and decided to try to solve it for myself before reading the answers, redrawing it clearly was the first thing I did.)
Others have already given some useful information about the solution, but I'd like to try to give you my intuition for how to approach it.
When I first saw the diagram, I was immediately suspicious that it was probably more symmetrical than it looked. You described it as a 'high school' problem; sometimes school exercises will involve doing gruntwork, but more often they will involve clever tricks you have to spot. And though the drawing obscures this, it's pretty easy to see that each terminal of the power supply is connected to exactly two resistors, each of which is connected to exactly two more resistors; and all the resistors in the circuit have the same value.
Symmetry is a critical principle to keep in mind, across all kinds of problems in all kinds of classes, and beyond that into the working world. You may never have to solve the current flow through a resistor network once you graduate, but you can amaze your coworkers all your life by making complex problems trivial using hidden symmetry. It often lets you prove that dramatic simplifications of a problem are possible with very little work.
In this case, as you can see in Trevor's beautiful drawing, the resistor network in this problem is extremely symmetrical. Trevor uses this to prove that the voltages in the middle nodes must be 12.5 V -- can you see why? Because, whatever the total resistance comes out to, between those nodes and V+/V-, it is obvious that it's the SAME above and below. It's then easy to see that no current can flow in resistors r6 and r9, since as Trevor has shown, the voltages across them are 0 by symmetry.
Now we can continue using the left-right and top-bottom symmetry of the whole diagram to see that the voltages must also be the same at all the mirror points on the left and the right, and the currents must be the same through the paired resistors r1-r3-r12-r14, r2-r7-r8-r13, and r4-r5-r10-r11. That alone reduces this to a fairly simple set of equations to solve.
But we can go a little further. Given two points with the same voltage, it won't change the current flows to connect them; and given a wire through which no current flows, it won't change anything to disconnect it. So we can make the diagram even more symmetric in a couple of ways, the easiest of which is this: Since we know no current flows left-right or right-left across the center (by symmetry), we can disconnect r4-r6-r10 from r5-r9-r11. This reduces the problem to a pair of parallel circuits, so you can solve one and then apply the result to both (and within each one, there are parallel strings of identical resistors where you can do the same thing again.)