@Sasha wrote above: "Is there a reference, or an argument, to demonstrate your claim about necessary condition for infinity divisibility in term of non-negativity of even-order cumulants?"
So I've taken a little bit of time to remind myself of this argument for a special case. Let $\{N_t : t \geq 0\}$ be a Poisson process, i.e. for each real $0, the random variables $N_{t_1},\ N_{t_2}-N_{t_1}, \ldots,N_{t_n} - N_{t_{n-1}}$ are independent and have Poisson distributions with expected values proportional to (and we may as well assume equal to) the lengths of the intervals $(0,t_1), (t_1,t_2),\dots$.
Now let $Y_t = \sum\limits_{n=1}^{N_t} X_n$ where $X_1,X_2,\dots$ are an i.i.d. sequence of real random variables. This is called a "compound Poisson process".
I claim every even-order cumulant of $Y_t$ is equal to $t$ times the corresponding even-order moment of $X_1$.
To see this, first recall some properties of cumulants: There's such a thing as the joint cumulant of any finite number of random variables: $\operatorname{cum}(A,B,C,\dots)$. If these are just $n$ copies of $A$ then $\operatorname{cum}(A,\dots,A)$ is what is called the $n$th cumulant $\operatorname{cum}_n(A)$ of $A$. The moment $E(ABCD)$ is a function of the cumulants that comes from enumerating the partitions of the set $\{A,B,C,D\}$ (similarly for sets of sizes other than 4, which is a convenient size to use here), thus: $ \begin{align} E(ABCD) & = \operatorname{cum}(A,B,C,D) \\ & {} + \underbrace{\operatorname{cum}(A,B,C)\operatorname{cum}(D) + \cdots\cdots}_{4\text{ terms}} \\ & {} + \underbrace{\operatorname{cum}(A,B)\operatorname{cum}(C,D)+ \cdots\cdots}_{\text{3 terms}} \\ & {} + \underbrace{\operatorname{cum}(A,B)\operatorname{cum}(C)\operatorname{cum}(D) + \cdots\cdots}_{6\text{ terms}} \\ & {} + \operatorname{cum}(A)\operatorname{cum}(B)\operatorname{cum}(C)\operatorname{cum}(D). \end{align} $ (This actually characterizes the cumulants completely if you do the similar thing for sets of sizes other than 4.) Constants come out, e.g. $\operatorname{cum}(3A,B,C,D) = 3\operatorname{cum}(A,B,C,D)$. A conditional cumulant $\operatorname{cum}(A,B,C,D \mid N=n)$ would be a function of $n$, and so $\operatorname{cum}(A,B,C,D \mid N)$ would be a random variable whose value is determined by that of $N$.
The law of total cumulance says $ \begin{align} \operatorname{cum}(A,B,C,D) & = \operatorname{cum}(\operatorname{cum}(A,B,C,D \mid N)) \\ & {} + \operatorname{cum}(\operatorname{cum}(A,B,C\mid N),\operatorname{cum}(D\mid N)) + \text{3 more terms} \\ & {} + \operatorname{cum}(\operatorname{cum}(A,B \mid N),\operatorname{cum}(C,D\mid N)) + \text{2 more terms} \\ & {} + \operatorname{cum}(\operatorname{cum}(A,B\mid N),\operatorname{cum}(C\mid N),\operatorname{cum}(D\mid N)) + \text{5 more terms} \\ & {} + \operatorname{cum}(\operatorname{cum}(A\mid N),\operatorname{cum}(B\mid N), \operatorname{cum}(C\mid N), \operatorname{cum}(D\mid N)), \end{align} $ and similarly for other sizes than 4.
Now apply this to $\operatorname{cum}_4\left(\sum\limits_{n=1}^{N_t} X_n\right)$.
In the course of this, repeatedly use the fact that $\operatorname{cum}_m\left( \sum\limits_{n=1}^{N_t} X_n \mid N_t \right) = N_t\operatorname{cum}_m(X_1)$. Then repeatedly use the fact that $\operatorname{cum}_{\ell}(N_t\operatorname{cum}_m(X_t)) = \operatorname{cum}_m(X_t) \cdot \operatorname{cum}_\ell(N_t)$ (the constant $\operatorname{cum}_m(X_t)$ was pulled out.) Repeatedly use the fact that $\operatorname{cum}_\ell(N_t) = t$ (regardless of the value of $\ell$).
You end up with $t$ times the sum of products of cumulants asserted above to add up to the 4th moment of $X_1$. Then apply to other sizes than 4. Even-numbered moments of $X_1$ are nonnegative; therefore, even-numbered cumulants of $Y_t$ are nonnegative.