There isn't even a solution if one drops the positivity requirement on $f$. Set \begin{equation} \mathcal S=\{\{0, 1\}, \{0, 2\}, \{0, 3\}, \{0, 4\}, \{1, 2\}, \{1, 3\}, \{1, 4\}, \{2, 3, 4\}\}. \end{equation} From the first $4$ members of $\mathcal S$ we get $f(1)=f(2)=f(3)=f(4)$. The $5$th, $6$th, and $7$th member yields $f(1)=f(2)=f(3)=f(4)=1/2$, so the last member yields the contradiction $1=f(2)+f(3)+f(4)=3/2$.

There isn't even a solution if one drops the positivity requirement on $f$. Set \begin{equation} \mathcal S=\{\{0, 1\}, \{0, 2\}, \{0, 3\}, \{0, 4\}, \{1, 2\}, \{1, 3, 4\}\}. \end{equation} From the first $4$ members of $\mathcal S$ we get $f(1)=f(2)=f(3)=f(4)$. Together with the $5$th member we get $f(1)=f(2)=f(3)=f(4)=1/2$, so the last member yields the contradiction $1=f(1)+f(3)+f(4)=3/2$.