Your link to Diophantine equations shows that you already have an idea of the answer.
The only actions available are filling a jug completely from the tub, emptying a jug completely, or pouring one jug into another. If we fill empty jug $k$ from the tub, then the amount of water we have is increased by $v_k$. Similarly, if we empty jug $k$ when it is full, we decrease the amount of water we have measured by $v_k$. If we empty a jug into another, the amount of water doesn't change. However, if jug $l$ is full, and we pour it into jug $m$, when $v_l>v_m$, then the amount of water left in jug $l$ is $v_l-v_m$, and emptying it will decrease the amount of water we have by that amount. From this, we can see that the only effect we can have on the total amount of water is to add or subtract $v_n$ repeatedly, for each $n\in[1,j]$.
This means that in order to get a total amount of water $n$ (you've used the same variable name for two things, tsktsk) there must be integers $z_1, z_2,\ldots z_j$ so that $z_1v_1+z_2v_2+\cdots+z_jv_j=n$. However, this is not a sufficient condition. For example, if your largest jug has volume $v_k$, and $n\gg v_k$, then you won't have anywhere to put the water you are measuring. The restrictions on the magnitude of $n$ for making a necessary and sufficient condition are tricky, so I'll get to that later. For now, let's suppose that we also have an infinite empty tub to put our water in.
Given volumes $v_1$ to $v_j$ and a goal $n$, the equation $z_1v_1+z_2v_2+\cdots+z_jv_j=n$, or $\sum_{k=1}^jz_kv_k=n$, is a Diophantine equation, because we are looking for integer solutions. So you could simply say that the problem is solvable when this equation has a solution. However, that isn't especially helpful, as it is just restating the question in an abstracted way.
Note that if $d$ is the smallest possible positive value of $\sum_{k=1}^jz_kv_k$, then $d=\gcd(v_1,v_2,\ldots,v_j)$, by Bézout's identity. Furthermore, every number that can be expressed this way (as an integer-multiple linear combination of $v$'s) is a multiple of $d$. Therefore, we have the requirement that $$\gcd(v_1,v_2,\ldots,v_j)\ \textrm{ divides }\ n.$$
If we assume the empty tub, this is a necessary and sufficient condition. It is especially nice because there is already an algorithm for computing the gcd, and the minimal coefficients of the corresponding linear combination: the extended Euclidean algorithm.
Given two integers $a$ and $b$, we can compute $\gcd(a,b)$ and integers $x$ and $y$ so that $ax+by=\gcd(a,b)$ as follows:
Let $r_0=a, r_1=b, s_0=1, s_1=0, t_0=0, $ and $t_1=1$. Then, for any index $i$, find $q_{i+1}, r_{i+1}$ so that $r_{i-1}=q_{i+1}r_i+r_{i+1}$ and $0\le r_{i+1}<r_i$. That is, $g_{i+1}=\left\lfloor\dfrac{r_{i-1}}{r_i}\right\rfloor$ and $r_{i+1}=r_{i-1}\mod r_i$. Also, let $s_{i+1}=s_{i-1}-q_{i+1}s_i$ and $t_{i+1}=t_{i-1}-q_{i+1}t_i$. When you reach $r_{i+1}=0$, stop. At this point, you have $\gcd(a,b)=r_i, x=s_i,$ and $y=t_i$.
This only finds the gcd of two numbers, however, $\gcd(a,b,c)=\gcd(\gcd(a,b),c)$, so you can apply this algorithm repeatedly, storing the coefficients at each stage. Also, you don't always need to test all of the jugs' volumes. As soon as you reach a point where $n$ is a multiple of the gcd you have found, you can stop.
(I will finish this answer later, including how to turn the linear combination back into a jug-pouring sequence.)
