
Let \( k \) be a positive real number. \( A \) and \( B \) play the following game: at the start, there are 80 zeroes arranged around a circle. Each turn, \( A \) increases some of these 80 numbers such that the total sum added is 1. Next, \( B \) selects ten consecutive numbers with the largest sum and reduces them all to 0. \( A \) wins the game if he/she can ensure that at least one of the numbers is \(\geq k\) at some finite point in time.

To determine all \( k \) such that \( A \) can always win the game, we analyze the strategy and the bounds involved.

First, we reduce the problem from \( F(80, 10) \) to \( F(8, 1) \). This reduction is valid because \( B \) can only select 10 consecutive buckets, and we can group these buckets into 8 larger buckets, each containing 10 of the original buckets. Thus, the problem simplifies to a scenario where \( B \) can only empty one of these larger buckets at a time.

We now play the game on \( F(8, 1) \). For \( B \), the optimal strategy is to empty the largest bucket all the time. Let the buckets be labeled \( 1 \) to \( 8 \) with amounts of water \( a_1, a_2, \ldots, a_8 \), arranged in decreasing order so that \( a_1 \geq a_2 \geq \ldots \geq a_8 \).

The key lemma is that after \( A \) makes a move, the sum of the largest \( k \) buckets is less than \( k \cdot \left(1 + \sum_{i=k}^{7} \frac{1}{i} \right) \). We prove this by downward induction.

For \( k = 8 \):
\[
\sum_{i=1}^{8} a_i \leq 1 + \sum_{i=2}^{8} a'_i \leq 1 + \frac{7}{8} \sum_{i=1}^{8} a'_i,
\]
where \( a'_i \) is the amount of water in the buckets in the previous round. Since initially \( \sum_{i=1}^{8} a_i = 1 \), it follows that \( \sum_{i=1}^{8} a_i < 8 \).

Assuming the statement is true for \( k = m \), for \( k = m-1 \):
\[
\sum_{i=1}^{m-1} a_i \leq 1 + \sum_{i=2}^{m} a'_i \leq 1 + \frac{m-1}{m} \sum_{i=1}^{m} a'_i < 1 + (m-1) \left(1 + \sum_{i=m}^{7} \frac{1}{i} \right).
\]
Thus, the statement holds for \( k = m-1 \) as well.

By induction, the statement is true for \( k = 1 \) as well, hence the largest bucket is less than \( 1 + 1 + \frac{1}{2} + \ldots + \frac{1}{7} \).

To show that this is the best attainable bound, \( A \)'s strategy is to ensure that \( a_1 = a_2 = \ldots = a_8 \) in each round. The amount of water in each bucket follows the sequence:
\[
\frac{1}{8}, \quad \frac{1}{8} \left(1 + \frac{7}{8} \right), \quad \frac{1}{8} \left(1 + \frac{7}{8} + \left(\frac{7}{8} \right)^2 \right), \ldots
\]
The limit of this sequence is 1. When the buckets are sufficiently close to 1, \( A \) will add \(\frac{1}{7}\) to the 7 largest buckets, \(\frac{1}{6}\) to the 6 largest buckets, and so on, finally adding 1 to the largest bucket left. The total amount of water in that bucket can be arbitrarily close to \( k = 1 + 1 + \frac{1}{2} + \ldots + \frac{1}{7} \).

Therefore, \( A \) can always win the game if \( k \leq 1 + 1 + \frac{1}{2} + \ldots + \frac{1}{7} \).

The answer is: \(\boxed{1 + 1 + \frac{1}{2} + \ldots + \frac{1}{7}}\).
Let $k$ be a positive real. $A$ and $B$ play the following game: at the start, there are $80$ zeroes arrange around a circle. Each turn, $A$ increases some of these $80$ numbers, such that the total sum added is $1$. Next, $B$ selects ten consecutive numbers with the largest sum, and reduces them all to $0$. $A$ then wins the game if he/she can ensure that at least one of the number is $\geq k$ at some finite point of time. 

Determine all $k$ such that $A$ can always win the game.