In the equation $2 b x+b=3 c x+c$, both $b$ and $c$ can take any of the values $1,2,3,4,5,6$. In how many cases will the solution of the equation be positive?

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By rearranging the equation, we get $(2 b-3 c) x=c-b$. The solution is positive if and only if $2 b-3 c$ and $c-b$ have the same sign, and neither is 0. This can happen in two ways:

1. $c-b>0$ and $2 b-3 c>0$.
2. $c-b<0$ and $2 b-3 c<0$.

For the first case, $c>b$ and $b>\frac{3}{2} c$, which implies $c>\frac{3}{2} c$. Therefore, $c$ must be negative, which is not possible since $c$ must be one of the values $1,2,3,4,5,6$. Thus, the first case cannot hold for any $b$ and $c$ values from the given set.

The second case holds if and only if

$$
c<b<\frac{3}{2} c
$$

Thus, we find the answer to the problem by counting how many pairs $(b, c)$ satisfy (*), where both $b$ and $c$ are one of the values $1,2,3,4,5,6$.

It is easy to verify that for $c=1, c=2$, and $c=6$, there is no integer between 1 and 6 that satisfies (*). In the other three cases, there is exactly one: if $c=3$, then $b=4$; if $c=4$, then $b=5$; and if $c=5$, then $b=6$.

Therefore, the solution to the equation will be positive in exactly three cases; the solutions in these cases are $1, \frac{1}{2}$, and $\frac{1}{3}$, respectively.