Patchouli is taking an exam with $k > 1$ parts, numbered Part $1, 2, \dots, k$. It is known that for $i = 1, 2, \dots, k$, Part $i$ contains $i$ multiple choice questions, each of which has $(i+1)$ answer choices. It is known that if she guesses randomly on every single question, the probability that she gets exactly one question correct is equal to $2018$ times the probability that she gets no questions correct. Compute the number of questions that are on the exam.

[i]Proposed by Yannick Yao[/i]
1. **Calculate the probability of getting no questions correct:**

   For each part \(i\) (where \(i = 1, 2, \dots, k\)), there are \(i\) questions, each with \(i+1\) answer choices. The probability of getting a single question wrong is \(\frac{i}{i+1}\). Therefore, the probability of getting all \(i\) questions wrong in part \(i\) is:
   \[
   \left( \frac{i}{i+1} \right)^i
   \]
   The probability of getting all questions wrong in all \(k\) parts is the product of the probabilities for each part:
   \[
   P(\text{no questions correct}) = \prod_{i=1}^k \left( \frac{i}{i+1} \right)^i
   \]

2. **Calculate the probability of getting exactly one question correct:**

   To get exactly one question correct, we consider the probability of getting one question correct in each part \(i\) and all other questions wrong. For part \(i\), the probability of getting one specific question correct and the remaining \(i-1\) questions wrong is:
   \[
   \left( \frac{i}{i+1} \right)^{i-1} \cdot \frac{1}{i+1}
   \]
   Since there are \(i\) questions in part \(i\), there are \(i\) ways to choose which question is correct. Thus, the probability for part \(i\) is:
   \[
   i \cdot \left( \frac{i}{i+1} \right)^{i-1} \cdot \frac{1}{i+1} = \left( \frac{i}{i+1} \right)^i
   \]
   This is the same as the probability of getting all questions wrong in part \(i\). Therefore, the total probability of getting exactly one question correct in any of the \(k\) parts is:
   \[
   P(\text{exactly one question correct}) = k \cdot \prod_{i=1}^k \left( \frac{i}{i+1} \right)^i
   \]

3. **Set up the given condition:**

   According to the problem, the probability of getting exactly one question correct is 2018 times the probability of getting no questions correct:
   \[
   k \cdot \prod_{i=1}^k \left( \frac{i}{i+1} \right)^i = 2018 \cdot \prod_{i=1}^k \left( \frac{i}{i+1} \right)^i
   \]
   Simplifying, we get:
   \[
   k = 2018
   \]

4. **Calculate the total number of questions on the exam:**

   The total number of questions is the sum of the number of questions in each part:
   \[
   1 + 2 + 3 + \dots + k = \frac{k(k+1)}{2}
   \]
   Substituting \(k = 2018\):
   \[
   \frac{2018 \cdot 2019}{2} = 2037171
   \]

The final answer is \(\boxed{2037171}\).