## Solution.

The answer is $7+\frac{1}{100}$.

Consider a collection of cups where 100 cups have a volume of 7, and 200 cups have a volume of $7+\frac{1}{100}$. We claim that this collection of cups satisfies the condition of the problem.

The total volume of the cups is $300 \cdot 7 + 200 \cdot \frac{1}{100} = 2102$.

For $n \in \{1, \ldots, 100\}$, we can take $n$ cups with a volume of 7, and their total volume is an integer.

For $n \in \{101, \ldots, 200\}$, we can take $n-100$ cups with a volume of 7 and 100 cups with a volume of $7+\frac{1}{100}$.

For $n \in \{201, \ldots, 300\}$, we can take $n-200$ cups with a volume of 7 and 200 cups with a volume of $7+\frac{1}{100}$.

Now assume that all cups have a volume less than $7+\frac{1}{100}$. Then the total volume of the 100 largest cups is less than 701. Take 100 cups whose total volume is an integer. Their total volume is less than the total volume of the 100 largest cups, so their volume is at most 700. We conclude that the total volume of the 100 smallest cups is also at most 700. However, then the total volume of the middle 100 cups must be greater than $2102 - 701 - 700 = 701$, which is a contradiction because it is greater than the total volume of the 100 largest cups.
## Task 2.

Masha has 300 cups. The volume of each cup is a positive real number. It is known that the sum of the volumes of all the cups is 2102, and for every $n \in\{1,2, \ldots, 300\}$, there exist $n$ cups whose total volume is a natural number. Determine the smallest possible volume of the largest cup.