By stacking the blocks so that the end of each one sits at a line drawn at the appropriate place on the block below, you can show that it is possible to stack them so that almost the entire top block extends past the end of the fourth block below it, and the entire block lies past the end of the bottom block by a noticeable distance.
The center of mass of an object is the weighted mean displacement of all of its mass points from some reference point, usually the origin of the coordinate system in which the object sits. This is the sum of the products of all the mass points with their distances from the origin, divided by the total mass of the object (rcm = Σmiri/M). If we set the center of mass of the object at the origin, we find that this sum equals zero. This tells us that if we exert a force on the object at its center of mass, this results in no torque on the object, but if we apply a force to the object at any other point, we exert a torque on the object about its center of mass.
The blocks in this demonstration are rectangular parallelepipeds of uniform density (at least roughly so). The sum above puts the center of mass of each block at the center of the block. If we could support one of the blocks at this point, then, there should be no torque, and the block should balance. We cannot do this without drilling through the block, but we can support the block along a line perpendicular to its long axis, underneath this point, and it should balance. We must be a bit careful, since any displacement of the center of mass to either side of this line produces a torque about the line of support. For the following discussion, we will assume that an individual block balances when we support it along this line.
If we wish the top block just to balance, then, we should be able to extend it so that the end of the block below is just at the aforementioned line of support. That is, if we call the length of the block l, then the center of mass of the top block is at l/2 from either end, and the end of the block below it sits at l/2 from its end; the top block sits so that half of it rests on the block below, and half overhangs it. To figure out where we must set these on a third block so that they just balance when they rest on it, we must find the center of mass of the first two together. The center of mass of the second block is at ml/2, where m is the mass of each block, and the center of mass of the top block is at ml. The center of mass of both together is [(ml/2) + ml]/(2m), or (3/4)l. So if we rest the top two blocks on a third block, the end of the third block must sit at (3/4)l; three-quarters of the second block must lie over the third block, and one-quarter can hang over the end. If we wish now to rest these three blocks on a fourth, we note that the center of mass of the top two is now at the end of the third, so that the center of mass of the three together is at [(ml/2) + 2ml]/(3m), or (5/6)l. The end of the fourth block must sit at (5/6)l; five-sixths of the third block must lie over the fourth block, and one-sixth can hang over the end. We find that for n blocks, relative to the end that is sitting on the stack of blocks below, the center of mass is at [(ml/2) + (n -1)l]/(nm), which we can write as [n - (1/2)]l/n, or (2n -1)l/(2n). So, calling the top block the first block and going down, the end of the second block must sit at l/2 along the first block, the end of the third block must sit at (3/4)l along the second block, the end of the fourth block must sit at (5/6)l along the third block, the end of the fifth block must sit at (7/8)l along the fourth block, and the end of the sixth (bottom, for our stack) block must sit at (9/10)l along the fifth block. (The blocks in this demonstration are actually numbered in the reverse order to this; number six is on top, and number 1 is on the bottom.)
Each block in the stack has a line drawn across it, close to the position that would put the center of mass of the blocks stacked above it at its end. When the blocks are stacked accordingly, as shown above, the stack just balances. If you were to move any of the blocks farther to the right, the blocks would no longer balance, and the stack would topple.
Now we come to the interesting part. If half the top block rests on the block below, the overhang of the top block relative to the second block is (1/2)l. Since three-quarters of the second block rests on the third block, its overhang relative to the third block is (1/4)l, and the overhang of the first block relative to the third block is [(1/2)l + (1/4)l] = (3/4)l. From the expression above for the center of mass, we see that the overhang for each block over the next is l - [n - (1/2)]l/n, or [1 - (n - (1/2))/n]l, or [1 - (2n -1)/2n]l, which reduces to [1/(2n)]l. The total overhang of the first block, then, over the end of n blocks below it, is the sum (1/2 + 1/4 + 1/6 + . . . + 1/(2n))l. So the top block overhangs the fourth block below it by (1/2 + 1/4 + 1/6 + 1/8)l, or (25/24)l (= 1.04l), and the bottom block (fifth block below it) by (1/2 + 1/4 + 1/6 + 1/8 + 1/10)l = (274/240)l, or 1.14l.
In the stack shown above, the overhangs are not quite as great as those listed above, but they are close. The left end of the top block is just inside the right end of the fourth block below it, almost even with it, and significantly past the right end of the bottom block.
References:
1) Resnick, Robert and Halliday, David. Physics, Part One, Third Edition (New York: John Wiley and Sons, 1977), pp. 162-5, 283-5.