The general topology will look like the following diagram. I've broken it into two parts -- left and right -- with dashed boxes. The left side is your switch at the bottom of the costume (as I understand where you want things) and the right side is the battery system at the top of the costume and flowing down from there (as you mentioned you could do.)

simulate this circuit – Schematic created using CircuitLab
As you can see, there are three wires which must go the entire distance: a NEGATIVE power wire, a POSITIVE power wire, and a SWITCHED POSITIVE power wire. You'll have to work out the details there. But you will need three conductors to make it down to the switch at the bottom of the costume.
(There are some other arrangements, but they all result in three wires because you say you want the battery supply on the opposite end of where the switch is at. If the battery and switch were together, then you could do with less than three.)
One thing to worry about is that a \$9\:\textrm{V}\$ battery doesn't actually provide full voltage for long. It rapidly drops downward. You could plan on as little as \$4.8\:\textrm{V}\$, in fact, if you had some kind of circuit that would keep on providing the right current and voltage to your LEDs. But that's extra complexity and cost and I'm not going there. So you need to figure out what you are willing to accept and how long you expect all this to run.
You don't say what kind of LEDs you are using for \$D_1\$ through \$D_{10}\$. If these are white LEDs and need about \$3.2-3.6\:\textrm{V}\$ each, then you may be in a heap of trouble. Suppose you planned on a full, perfect \$9\:\textrm{V}\$ battery supply (never will happen; not even close) and decided to use \$R=\frac{9\:\textrm{V}-2\cdot 3.5\:\textrm{V}}{20\:\textrm{mA}}=100\:\Omega\$ as your current limiting resistor. Well, the battery voltage will quickly drop towards \$8\:\textrm{V}\$ (probably within a half-hour or so) and you will be supplying only \$10\:\textrm{mA}\$ (perhaps slightly more) then. Already, they are dimming down. By the time the battery gets down to \$7\:\textrm{V}\$ (where there is still some life left, too) your LEDs will be barely visible.
This is a serious difficulty in using a \$9\:\textrm{V}\$ alkaline battery. They don't hold their voltage for long. A circuit designed well for one of these should be able to operate down to some much lower voltage. At least down to \$7\:\textrm{V}\$ and perhaps even lower, towards \$6\:\textrm{V}\$.
But if pairing the LEDs will require all of that (and more, perhaps), then it is very difficult to make this work well with nothing more than a dropping resistor for each pair of LEDs.
So. Disclose your exact LEDs (we need to know operating current and voltage.) Disclose how long you want all this to operate (this has a great deal to do with the battery supply itself as well as possible circuitry options to consider.) Disclose how tolerant you are willing to be about the LEDs dimming down over the period of operation you need.
The value for the resistors, including \$R_1\$, may depend upon all of the above. And, depending upon the details if and when you answer, the resistor idea may not even work well at all making the entire question moot.
Thanks for the link. They say \$3.3\:\textrm{V}\$ provides about \$5\:\textrm{mA}\$ with \$100\:\Omega\$ in series. So that's \$2.8\:\textrm{V}\$, at that current. They also say they can operate up to \$6\:\textrm{V}\$, so in very round numbers I'd say you should plan \$3.0\:\textrm{V}\$ each (ignoring the resistor, assuming you are willing to bypass it -- short it out.) If you keep the \$100\:\Omega\$ resistor in place and want to operate these at full brightness, I'd say you should plan on about \$4.8-5.0\:\textrm{V}\$ each (which means you won't be putting them in series pairs.)
So. You should do some testing.
I'd recommend that you start out by considering the idea of bypassing the \$100\:\Omega\$ resistor that is included, as that may permit you to still use them in pairs. If you try that, consider putting two of them in series and bypassing (shorting out) just one of the two resistors in the pair. See how that works in terms of brightness and, if 10 of them running that way, longevity. Also set \$R_1=220\:\Omega\$, or thereabouts. Up or down, but in that neighborhood to start.
If that works for you, then go with that design. If not, you'll need to consider alternative ideas.