Put your money where your mouth is

This problem is best solved using elimination grids . Let's start by drawing out the grid of the numbers versus the roman numerals:

$\begin{array} {| c | c | c | c | } \hline \text{Roman\ Number} & 1 & 2 & 3 \\ \hline \rm{I} & & & \\ \hline \rm{II} & & & \\ \hline \rm{III} & & & \\ \hline \end{array}$

From the first and third points:

The person whose favorite natural number is 1 don't like the roman letter $\rm{III}$ , and

The person whose favorite roman number is $\rm{II}$ don't like the number 3.

Adding these two points into the grid gives us

$\begin{array} {| c | c | c | c | } \hline \text{Roman\ Number} & 1 & 2 & 3 \\ \hline \rm{I} & & & \\ \hline \rm{II} & & & \times \\ \hline \rm{III} & \times & & \\ \hline \end{array}$

From the second point: "Exactly one of these 3 people have the same favorite roman numeral and natural number."

So either one of these must be true only:
Case one : The person who likes the roman numeral $\rm{I}$ also likes the natural number 1.
Case two : The person who likes the roman numeral $\rm{II}$ also likes the natural number 2.
Case three : The person who likes the roman numeral $\rm{III}$ also likes the natural number 3.

Suppose it's case one, then updating the grid gives us:

$\begin{array} {| c | c | c | c | } \hline \text{Roman\ Number} & 1 & 2 & 3 \\ \hline \rm{I} & \checkmark & \times & \times \\ \hline \rm{II} & \times & & \times \\ \hline \rm{III} & \times & & \\ \hline \end{array} \quad\Rightarrow \quad \begin{array} {| c | c | c | c | } \hline \text{Roman\ Number} & 1 & 2 & 3 \\ \hline \rm{I} & \checkmark & \times & \times \\ \hline \rm{II} & \times & \checkmark & \times \\ \hline \rm{III} & \times & \times & \checkmark \\ \hline \end{array}$

However, this tells us that the person who likes the roman numeral $\rm{II}$ also likes the number 2, which is a contradiction. Thus case one is not allowed.

Similarly, we can work our way to show that case two can't be true as well:

$\begin{array} {| c | c | c | c | } \hline \text{Roman\ Number} & 1 & 2 & 3 \\ \hline \rm{I} & & \times & \\ \hline \rm{II} & \times & \checkmark & \times \\ \hline \rm{III} & \times & \times & \\ \hline \end{array} \quad\Rightarrow \quad \begin{array} {| c | c | c | c | } \hline \text{Roman\ Number} & 1 & 2 & 3 \\ \hline \rm{I} & \checkmark & \times & \times \\ \hline \rm{II} & \times & \checkmark & \times \\ \hline \rm{III} & \times & \times & \checkmark \\ \hline \end{array}$

What's left to confirm is to check wheter case three works or not, and the illustration in the grid below does agree with all the conditions.

$\begin{array} {| c | c | c | c | } \hline \text{Roman\ Number} & 1 & 2 & 3 \\ \hline \rm{I} & & & \times \\ \hline \rm{II} & & & \times \\ \hline \rm{III} & \times & \times & \checkmark \\ \hline \end{array} \quad\Rightarrow \quad \begin{array} {| c | c | c | c | } \hline \text{Roman\ Number} & 1 & 2 & 3 \\ \hline \rm{I} & \times & \checkmark & \times \\ \hline \rm{II} & \checkmark & \times & \times \\ \hline \rm{III} & \times & \times & \checkmark \\ \hline \end{array}$

From here, we can improve the second point to: Exactly one of these 3 people have the favorite roman numeral of $\rm{III}$ and favorite natural number of 3. And because everyone are all perfectly rational, they are able to figure out this very fact as well. And because Taylor made the claim that he still can't solve it, this means that he is in the position where there no other informations given to him even after knowing that exactly one of these 3 people have the favorite roman numeral of $\rm{III}$ and favorite natural number of 3. This can only mean that Taylor himself whose favorite numbers are both these numbers. Constructing the grids for the people versus the favorite numbers gives us:

$\begin{array} {| c | c | c | c | } \hline \text{Name\ Number} & 1 & 2 & 3 \\ \hline \text{Ellinor} & & & \times \\ \hline \text{Taylor} & \times & \times & \checkmark \\ \hline \text{McGregor} & & & \times \\ \hline \end{array}$

And because everyone knows their own favorite number, (Andrew) Ellinor would have known his own number and also Taylor's number, thus he is able to complete the grid (as shown below).

$\begin{array} {| c | c | c | c | } \hline \text{Name\ Number} & 1 & 2 & 3 \\ \hline \text{Ellinor} & \checkmark & \times & \times \\ \hline \text{Taylor} & \times & \times & \checkmark \\ \hline \text{McGregor} & \times & \checkmark & \times \\ \hline \end{array} \quad \text{or} \quad \begin{array} {| c | c | c | c | } \hline \text{Name\ Number} & 1 & 2 & 3 \\ \hline \text{Ellinor} & \times & \checkmark & \times \\ \hline \text{Taylor} & \times & \times & \checkmark \\ \hline \text{McGregor} & \checkmark & \times & \times \\ \hline \end{array}$

Regardless of whether Ellinor's favorite number is 1 or 2, he is able to figure out Taylor's 2 favorite number, and thus is able to figure out the remaining possible favorite numbers from McGregor. Hence, the answer is indeed $\boxed{\rm{III}}$ .