On day $1$ of the new year, John Adams and Samuel Adams each drink one gallon of tea. For each positive integer $n$, on the $n$th day of the year, John drinks $n$ gallons of tea and Samuel drinks $n^2$ gallons of tea. After how many days does the combined tea intake of John and Samuel that year first exceed $900$ gallons?

[i]Proposed by Aidan Duncan[/i]

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[i]Solution. [/i] $\boxed{13}$
The total amount that John and Samuel have drank by day $n$ is 
$$\dfrac{n(n+1)(2n+1)}{6}+\dfrac{n(n+1)}{2}=\dfrac{n(n+1)(n+2)}{3}.$$ 
Now, note that ourdesired number of days should be a bit below $\sqrt[3]{2700}$. Testing a few values gives $\boxed{13}$ as our answer.
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To determine after how many days the combined tea intake of John and Samuel first exceeds 900 gallons, we need to calculate the total amount of tea they drink over a period of \( n \) days.

1. **Calculate John's total tea intake:**
   John drinks \( n \) gallons of tea on the \( n \)-th day. Therefore, the total amount of tea John drinks over \( n \) days is the sum of the first \( n \) positive integers:
   \[
   \sum_{k=1}^n k = \frac{n(n+1)}{2}
   \]

2. **Calculate Samuel's total tea intake:**
   Samuel drinks \( n^2 \) gallons of tea on the \( n \)-th day. Therefore, the total amount of tea Samuel drinks over \( n \) days is the sum of the squares of the first \( n \) positive integers:
   \[
   \sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}
   \]

3. **Combine the total tea intake:**
   The combined total tea intake of John and Samuel over \( n \) days is:
   \[
   \text{Total tea intake} = \sum_{k=1}^n k + \sum_{k=1}^n k^2 = \frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6}
   \]

4. **Simplify the expression:**
   To combine these fractions, we need a common denominator:
   \[
   \frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6} = \frac{3n(n+1)}{6} + \frac{n(n+1)(2n+1)}{6} = \frac{n(n+1)(3 + 2n+1)}{6} = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3}
   \]

5. **Set up the inequality:**
   We need to find the smallest \( n \) such that the combined tea intake exceeds 900 gallons:
   \[
   \frac{n(n+1)(n+2)}{3} > 900
   \]

6. **Solve the inequality:**
   Multiply both sides by 3 to clear the fraction:
   \[
   n(n+1)(n+2) > 2700
   \]

   To estimate \( n \), we can approximate \( n^3 \approx 2700 \):
   \[
   n \approx \sqrt[3]{2700} \approx 13.98
   \]

   We test \( n = 13 \) and \( n = 14 \):

   For \( n = 13 \):
   \[
   13 \cdot 14 \cdot 15 = 2730
   \]

   For \( n = 12 \):
   \[
   12 \cdot 13 \cdot 14 = 2184
   \]

   Since \( 2184 < 2700 \) and \( 2730 > 2700 \), the smallest \( n \) for which the combined tea intake exceeds 900 gallons is \( n = 13 \).

The final answer is \(\boxed{13}\)