Solution:

If 39 thieves take 101 euro each and the last one takes 61 euro, then the only poor groups are those having as a member the last thief. So this distribution of the money gives $\binom{39}{4}$ poor groups. We shall prove that this is the required minimum.

Let $r$ be the number of all possible divisions of the thieves in 8 groups of five thieves each. Any such division has at least one poor group. The total number of the groups in all divisions is $8r$. Each group takes part in $8r / \binom{40}{5}$ divisions. Therefore each poor group is counted exactly $8r / \binom{40}{5}$ times and this gives at least $r$. Therefore we have at least $r \binom{40}{5} / 8r = \binom{39}{4}$ poor groups.