In the land of Chaina, people pay each other in the form of links from chains. Fiona, originating from Chaina, has an open chain with $2018$ links. In order to pay for things, she decides to break up the chain by choosing a number of links and cutting them out one by one, each time creating $2$ or $3$ new chains. For example, if she cuts the $1111$th link out of her chain first, then she will have $3$ chains, of lengths $1110$, $1$, and $907$. What is the least number of links she needs to remove in order to be able to pay for anything costing from $1$ to $2018$ links using some combination of her chains? 

[i]2018 CCA Math Bonanza Individual Round #10[/i]
1. To solve this problem, we need to determine the minimum number of links Fiona must cut to be able to pay for any amount from 1 to 2018 links using the resulting chains. The key insight is that any number can be represented as a sum of powers of 2. This is because every integer can be uniquely represented in binary form, which is a sum of distinct powers of 2.

2. First, let's list the powers of 2 that are less than or equal to 2018:
   \[
   2^0 = 1, \quad 2^1 = 2, \quad 2^2 = 4, \quad 2^3 = 8, \quad 2^4 = 16, \quad 2^5 = 32, \quad 2^6 = 64, \quad 2^7 = 128, \quad 2^8 = 256, \quad 2^9 = 512, \quad 2^{10} = 1024, \quad 2^{11} = 2048
   \]
   Since \(2^{11} = 2048\) is greater than 2018, we do not need it. Therefore, the relevant powers of 2 are:
   \[
   1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024
   \]

3. The number of these powers of 2 is 11. However, we need to consider the cuts required to obtain these lengths. Each cut creates additional chains. For example, if we cut a chain of length 2018 at the 1024th link, we get two chains: one of length 1024 and another of length 994. We continue this process to obtain all the required lengths.

4. To minimize the number of cuts, we should always cut the largest possible chain first. Here is the step-by-step process:
   - Cut at 1024: Chains of lengths 1024 and 994.
   - Cut 994 at 512: Chains of lengths 1024, 512, and 482.
   - Cut 482 at 256: Chains of lengths 1024, 512, 256, and 226.
   - Cut 226 at 128: Chains of lengths 1024, 512, 256, 128, and 98.
   - Cut 98 at 64: Chains of lengths 1024, 512, 256, 128, 64, and 34.
   - Cut 34 at 32: Chains of lengths 1024, 512, 256, 128, 64, 32, and 2.
   - Cut 2 at 2: Chains of lengths 1024, 512, 256, 128, 64, 32, 2, and 1.

5. We have now obtained all the required chain lengths: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, and 1024. The number of cuts made is 10.

The final answer is \(\boxed{10}\).