General problem

For every positive integer $n$ , Cape Town Bank issues some coins that has $\frac{1}{n}$ value. Let a collection of such finite coins (coins does not neccesarily have different values) which sum of their value is less than $r-\frac{1}{2}$ . Prove that we can divide the collection into at most $r$ groups such that sum of all coins' value does not exceed $1$ .

Proof with induction

For $r=1$ it is trivial.

Now assume it is proven for $r = 1,\cdots, R-1$ .
We will prove it now for $r=R$ .

If there is a sum of some such reciprocals equal to $1$ , we just take these in one group and by induction it will work for the other coins.

So, if it would be impossible, we can look to the coins with value equal to a fraction of

$S_1:$ $1, \frac{1}{2}, \frac{1}{2^2} \cdots$
$S_2:$ $\frac 13, \frac{1}{2*3}, \frac{1}{2^2*3} \cdots$
$\cdots$

$S_{R}:$ $\frac{1}{2R-1}, \frac{1}{2(2R-1)},\frac{1}{2^2(2R-1)}\cdots$
$\cdots$ (others)

CLAIM
The sum of coins in $S_i$ for $1 \le i \le R$ is each time smaller than one.
Proof: Else, the sum of som coins would be equal to one.

Now for each coin with an other value, we place it in a $S_i$ with $i \le R$ such that the sum of values in $S_i$ keeps being smaller or equal to one.

This is possible, as $min \{ S_i | i \le R\} \le \frac{R-0.5}{R} = 1 - \frac{1}{2R}$ and each new value is maximum $\frac{1}{2R+1}$ .
Hence we can place all others.