Let $B$ be a Boolean Algebra.
A strictly positive measure on $B$ is a function $m$ from $B$ to $[0,1]$ such that (i) $m(b)=0$ iff $b=0$, (ii) $m(1)=1$, and (iii) $m(a+b)=m(a)+m(b)$ whenever $a$ and $b$ are disjoint.
Is there a strictly positive measure on every countable Boolean Algebra?
Yes. Let $M$ be the space of all measures on $B$. This is a compact space when endowed with with the pointwise convergence topology since it is a closed subspace of $[0,1]^B$. If $b$ is a nonzero element of $B$, then the set $U_b = \lbrace m \in M : m(b) \gt 0 \rbrace$ is open and dense in $M$. By the Baire Category Theorem, the intersection of all these sets is nonempty.
For an explicit construction, let $b_1,b_2,\ldots$ enumerate $B\setminus\lbrace0\rbrace$ and for each $n$ let $m_n$ be a measure on $B$ such that $m_n(b_n) \gt 0$ (e.g. let $m_n$ be the characteristic function of an ultrafilter containing $b_n$). Then $m = \sum_{n=1}^\infty 2^{-n}m_n$ is as required.
