Solution:
We can take all $\{a_i\}$ to lie in the range $(0,1)$ and all $\{b_i\}$ to be $0$ or $1$. The largest positive value of the sum of $(a_i - b_i)$ for any subset is achieved by taking the subset of those $i$ for which $b_i = 0$. Similarly, the largest negative value is achieved by taking those $i$ for which $b_i = 1$. So the worst subset will be one of those two.

If $a_i < a_j$, then we cannot have $b_i = 1$ and $b_j = 0$ if the set of $b_i$ is to minimise the maximum sum, because swapping them would reduce the sum of $a$'s with $b = 0$ and the sum of $(1 - a)$'s with $b = 1$. So if we order the $a$'s so that $a_1 \leq a_2 \leq \ldots \leq a_n$, then a best set of $b$'s is $b_i = 0$ for $i \leq$ some $k$, and $b_i = 1$ for $i > k$. [If some of the $a_i$ are equal, then we can find equally good sets of $b$'s that do not have this form, but we cannot get a lower maximum sum by departing from this form.]

Let $L_i = a_1 + a_2 + \ldots + a_i$, and $R_i = a_{i + 1} + a_{i + 2} + \ldots + a_n$. As we increase $i$ the sums $L_i$ increase and the sums $R_i$ decrease, so for some $k$ we must have $L_k < R_k$, $L_{k + 1} \geq R_{k + 1}$. Either $k$ or $k + 1$ must correspond to the optimum choice of $b$'s to minimise the maximum sum.

Now assume that the $a$'s form a maximal set, in other words they are chosen so that the minimum is as large as possible. We show first that in this case $L_{k + 1} = R_k$. Suppose $L_{k + 1} < R_k$. Then we could increase each of $a_{k + 1}, a_{k + 2}, \ldots, a_n$ by $\varepsilon$. This would leave $L_k$ unaffected, but slightly increase $L_{k + 1}$ and slightly reduce $R_k$. For small $\varepsilon$ this does not change the value of $k$, but increases the smaller of $L_{k + 1}$ and $R_k$, thus increasing the minimum and contradicting the maximality of the original $a$'s. Similarly, if $L_{k + 1} > R_k$, we could decrease each of $a_1, a_2, \ldots, a_{k + 1}$ by $\varepsilon$, thus slightly increasing $R_k$ and reducing $L_{k + 1}$.

Suppose not all of $a_1, a_2, \ldots, a_{k + 1}$ are equal. Take $i$ so that $a_i < a_{i + 1}$. Now increase each of $a_1, a_2, \ldots, a_i$ by $\varepsilon$ and reduce each of $a_{i + 1}, a_{i + 2}, \ldots, a_{k + 1}$ by $\varepsilon'$, with $\varepsilon$ and $\varepsilon'$ sufficiently small that we do not upset the ordering or change the value of $k$, and with their relative sizes chosen so that $L_{k + 1}$ is increased. $R_k$ is also increased, so we contradict the maximality of the $a$'s. Hence all $a_1, a_2, \ldots, a_{k + 1}$ are equal. Similarly, we show that all of $a_{k + 1}, \ldots, a_n$ are equal. For if not we can increase slightly $a_{k + 1}, \ldots, a_j$ and reduce slightly $a_{j + 1}, \ldots, a_n$ to get a contradiction.

So we have established that all the $a$'s must be equal. Suppose $n$ is odd $= 2m + 1$ and that all the $a$'s equal $x$. Then for the optimum $k$ we have $(k + 1)x = (2m + 1 - k)(1 - x)$, hence $k + 1 = (2m + 2)(1 - x)$ and the maximum difference is $(k + 1)x = (2m + 2)(1 - x)x$. This is maximised by taking $x = 1/2$, $k = m$, and is $(m + 1)/2 = (n + 1)/4$.

If $n$ is even $= 2m$, then for the optimum $k$ we have $(k + 1)x = (2m - k)(1 - x)$, so $k + 1 = (2m + 1)(1 - x)$, and the maximum difference is $(k + 1)x = (2m + 1)(1 - x)x$. However, in this case we cannot take $x = 1/2$, because that would give $k = m - 1/2$ which is non-integral, so we take $k = m - 1$ or $m$, both of which give a maximum difference of $m(m + 1)/(2m + 1) = n(n + 2)/(4n + 4) < (n + 1)/4$.