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: $\boxed{1 + 1 + \frac{1}{2} + \ldots + \frac{1}{7}}$.