Yes, but Image may be NSFW.
Clik here to view.
I (and I expect most experienced folks) tend to forget that newcomers don't have the other laws that apply hard wired in to their brains. In this case kirkoff's law comes in to effect because this is a resistor network. We need a bit more complete (but still not anywhere near reality such as a spice model) drawing of this equivalent circuit. We still have the external voltage divider R1 / R2 but the cpu input pin is a bit more detailed, specifically its input resistance (which is probably really higher than a meg) and the substrate diode which is the real issue in over voltage on the input pin. This drawing contains a more accurate representation of both the voltage divider and the 20 feet of 26 wire to the sensors (which we will use later).
For why the voltage divider works the way it does, we need to add kirkoff's law to ohm's law. Essentially kirkoffs law says the sum of the currents flowing in to a node equals the sum of the currents flowing out of the node. For us that means resistors in series add their values (so the 100 ohm and 68 ohm resistor provide 168 ohms across 5V). The current flowing in the 1M gate resistance provides so little current that it can be ignored (if it was 100 ohms the the cpu input pin would see 50 ohms rather than 100 and the voltage would change in an undesirable way). Essentially the values (under normal conditions) of the components in the input pin for the cpu make have no effect on the voltage divider and the voltage at the input pin will be 3 volts. The problem with supplying more than 3.3 (actually more than 3.9 or so) volts to the input pin is the substrate diode that connects from the pin to the 3.3V supply in the chip. If we supply 5V to the input from a low resistance source (such as a 5V output driver that can source 40ma) the diode will conduct and attempt (and if it has enough current will succeed) to drag the 3.3V power supply and the chip to 5V. High current and damage to the 3.3V chip are likely to occur here. I know this from unfortunate experience because I tried to do some 3.3V level translation with a 74hc series chip (with the mistaken belief that it didn't have the substrate diode, which in fact is only true for some members of the 74ahc family). Luckily I had disconnected the 3.3V device that I was driving before trying this so all I blew was the 74hc device and the driver pin on my arduino CPU chip as it dragged the 3.3V regulator up to about 4.5 volts through the 74hc chip.
A 74ahc chip in the same circuit works fine (and the document that stated 74hc did too has been removedfrom the site where I found it Image may be NSFW.
Clik here to view. ). With that done, now on to your sensors 20 feet over 26 wire ... To the left in the drawing you see an again simplified representation of what that wire looks like electrically. It has resistance (about .8 ohm for 20 feet of 26 according to an online calculator) and inductance and capacitance. The resistance might affect the voltages seen at the sensors, it is also small enough it may not matter. There are two considerations here: first as noted the sensors want somewhere close to 5V (or 3.3V) for power and may malfunction if the power drops below about 4.5 or 3V respectively. Bypass capacitors at the sensor end will help with this (as will the fact that the sensors don't draw much current typically. If you can using two wires to supply 5V and ground to the sensor end of the cable and then with bypass capacitors on the sensor end of the 5V lines use a 3.3 V LDO regulator to provide the 3.3V to the sensors. The other related issue to be aware of is signal level noise immunity. A digital input typically accepts between 0v and .7V as being low and 2.4V of higher as a high (and voltages between .71 and 2.3 are undesirable for a variety of reasons, and should be transitioned through fast). The capacitance, inductance and resistance in the 20 feet of wire conspire to make this difficult. The effect of the three is to slow down the transition from a 1 to a 0 (and visa versa). If there is sufficient current flowing (which there likely isn't in this case), the resistance can produce a voltage drop in the wire where 0V at the sensor shows up as .8V at the cpu end (the .8v being dropped by the .8 ohm resistance of the wire if there was 1 amp of current flowing). The effect this can produce is similar to what bypass capacitors help with on the power supply leads: it works 99% of the time but %1 fails because a 0 was read as a 1 due to voltage drop in the cable. It would be a good bet to try the sensors on the cable preferably in the place where it will be used to make sure it works acceptably. If not you would need to consider line driver options to get the data reliably from the sensors to the cpu (although I suspect this may work fine as is).
As it stands you won't get plated through holes. Your pcb is set to single sided and it needs to be double sided to get the top copper hole to plate through to. If you have no objection to double sided boards, that would fix your jumper issue as well. If you change to a double sided board on both you will get the top side pads plated through holes and on the cpu module can route the jumper connections on the top layer getting rid of the jumpers (I have been assuming you wanted to keep the design single sided, which perhaps you do).
Peter