Nobody ?

Anyone ?

The maximum is $10^6$ This can be achieved by having $\frac{1}{1000}$ in all squares except the top 2 by 3, and then having $10^1-\frac{94}{10000}$, $10^2-\frac{94}{10000}$, ..., $10^6-\frac{94}{10000}$ in a cyclic order, and have the frogs jump around in a circle. We shall now prove $10^7$ is unattainable. In the increase from $10^1$ to $10^2$ there is an increase of 90. As 5 squares were revealed and 5 taken away, at least one of the revealed squares must have a value strictly greater than $\frac{90}{6} = 15$. However as the total value after is only 100, the square must have value at most 100. Therefore the square has value in the range $(15,100)$. Similarly for the other jumps up to $10^6$ to $10^7$ we must have squares in the ranges: $(1.5*10^2,10*10^2)$, $(1.5*10^3,10*10^3)$, ..., $(1.5*10^6, 10*10^6)$. These 6 squares are in non-intersecting ranges, so they must all be unique. However as there are only 5 frogs, at least one of the squares must be visible at all times, which is impossible as even the smallest possible value, 15, creates too big of a sum for the first position. Therefore, Kostya cannot get a sum of $10^7$, so the maximal value is $\boxed{10^6}$.

I have a question: Why at least one of the revealed squares must have a value strictly greater than $\frac{90}{6} = 15$ and not $\frac{90}{5} = 18$ ?