the monster is a nice example of how the so-called rigidity method for the inverse Galois problem works. There is a lot of beautiful mathematics behind this, I will sketch the different steps.
A general remark: it is known that every profinite group, i.e. every group which could be a Galois group of some field extension, is indeed the Galois group of some Galois extension. This is still quite elementary (a result of Leptin, also proved by Waterhouse). To make the inverse Galois question more interesting, we should therefore consider a fixed base field $K$. We can't hope that any profinite group will still be a Galois group over $K$ - every Galois group over $K$ is a quotient of the absolute Galois group of $K$, and therefore the cardinality of a Galois group over $K$ is bounded from above, whereas it is easy to see that there are profinite groups which are "strictly bigger" in cardinality. So a very reasonable question is indeed to ask whether every finite group is a Galois group over some fixed base field $K$. The most natural case is to ask the question for $K = \mathbb{Q}$, but also other base fields can be considered - for example, for $K = \mathbb{C}(t)$ the inverse Galois conjecture is true. In fact, the inverse Galois problems for different base fields $K$ are sometimes closely linked; the method which I will sketch below is a perfect illustration for this, since we will have the consider four different base fields: $\mathbb{C}(t)$, $\overline{\mathbb{Q}}(t)$, $\mathbb{Q}(t)$, and of course $\mathbb{Q}$.
(1) Start with the fact that each finite group $G$ can be realized as a Galois group over $\mathbb{C}(t)$. This follows from the theory of coverings of Riemann-surfaces; if $G$ can be generated by $n - 1$ elements, then we can realize $G$ as a quotient of the fundamental group of the punctured Riemann sphere $\pi_1^{\text{top}}(\mathbb{P}^1(\mathbb{C}) \setminus \{P_1,P_2,\,\cdots,P_n\})$, where we choose the points $P_1,P_2,\,\cdots,P_n$ to be rational.
(2) We use the theory of the étale fundamental group to get an isomorphism $\pi_1(\mathbb{P}^1(\mathbb{C}) \setminus \{P_1,P_2,\,\cdots,P_n\}) = \pi_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P_1,P_2,\,\cdots,P_n\})$, which allows us to realize $G$ as a Galois group over $\overline{\mathbb{Q}}(t)$. [$\pi_1$ is the étale fundamental group - here the profinite completion of the topological version.]
[Of course, this is already advanced material; see the Wikipedia article for background. For a proper introduction to the theory, there is SGA 1 by Grothendieck; and the recent book "Galois groups and fundamental groups" by Tamas Szamuely is a very gentle introduction (and does all this in detail).]
(3) There exists an exact sequence
$1 \to \pi_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P_1,P_2,\,\cdots,P_n\}) \to \pi_1(\mathbb{P}^1(\mathbb{Q}) \setminus \{P_1,P_2,\,\cdots,P_n\}) \to \text{Gal}(\overline{\mathbb{Q}}|\mathbb{Q}) \to 1$
(this is a very fundamental result; see again the books I mentioned) and basically we now want to extend a surjective homomorphism from $\pi_1(\mathbb{P}^1(\overline{\mathbb{Q}}) \setminus \{P_1,P_2,\,\cdots,P_n\})$ to $G$ to a surjective homomorphism from $\pi_1(\mathbb{P}^1(\mathbb{Q}) \setminus \{P_1,P_2,\,\cdots,P_n\})$ to $G$. Of course this depends heavily on the structure of the group $G$. This works for many finite simple groups; the construction is quite general, but for particular groups there is always some technical work to do to show that the method applies. In particular it works for finite groups with a trivial centre, and a rigid system of rational conjugacy classes - these are quite technical conditions, of course, and I will just state the definitions. An $n$-tuple of conjugacy classes $C_1,C_2,\,\cdots,C_n$ of $G$ is rigid if there exists $(g_1,g_2,\,\cdots,g_n) \in G^n$ such that the $g_i$ generate $G$, $g_1g_2\cdots g_n = 1$ and $g_i \in C_i$, and if moreover $G$ acts transitively on the set of all such $n$-tuples $(g_1,g_2,\,\cdots,g_n)$. A conjugacy class $C$ of $G$ is rational if $g \in C$ implies $g^m \in C$ for all $m$ coprime to the order of $G$. I won't explain why precisely these conditions give you what you want, since it is really technical. Szamuely explains this very clearly. The conditions can be generalized, but that doesn't make it more readable...
[References: section 4.8 in Szamuely's book I mentioned above, and also Serre's wonderful book "Topics in Galois theory", which should maybe be called "Topics in inverse Galois theory" :)]
(4) The previous step allows us to descend from $\overline{\mathbb{Q}}(t)$ to $\mathbb{Q}(t)$, i.e. to realize $G$ as a Galois group of a regular extension - another technical notion which I won't explain, but it is not unimportant - of $\mathbb{Q}(t)$. To descend from $\mathbb{Q}(t)$ to $\mathbb{Q}$, there is Hilbert's irreducibility theorem, or some slight generalization (I don't remember exactly).
According to Thompson, the Monster has a rigid system of three rational conjugacy classes of orders 2, 3 and 29. So the method will apply; of course, I guess that it will be very hard to construct these conjugacy classes, and it is clear that the classification of finite simple groups has played a very big role in these developments. (But I am not a group theorist, so anyone who knows how this works is welcome to give additional information about this construction :))
So I hope this gives you an idea; I wrote this up in a hurry, so suggestions to make this clearer or more coherent (or of course corrections of details which I got wrong) are always welcome.
Best Answer
There is a version of the Chebotarev density theorem for finitely generated fields, or more precisely, after spreading out, for an étale Galois cover of schemes of finite type over a ring of $S$-integers. This is a consequence of work surrounding the Weil conjectures. See Lemma 1.2 in Torsten Ekedahl, An effective version of Hilbert's irreducibility theorem, Séminaire de Théorie des Nombres, Paris 1988-89, Birkhäuser 1990.
But what I would recommend trying first is to use the monodromy groups. If you spread out your Galois extension to a finite branched cover of a dense open subscheme $U$ of $\mathbf{P}^n_{\mathbf{Q}}$, then the monodromy group (inertia group) associated to any irreducible divisor on $U$ is contained in your Galois group. In particular, if there is a divisor that ramifies in the simplest possible way (ramification indices $2,1,1,\ldots,1$), then you know that your group contains a transposition.
The other approach to try, already mentioned by others in the comments, is specialization, which amounts to spreading out as above and then restricting to the cover above an irreducible closed subscheme. More concretely, you could plug in rational numbers for some or all of $y_1,\ldots,y_n$ such that you get a separable polynomial (of the same degree as the original polynomial). Or you could spread out to a scheme of finite type over $\mathbf{Z}$ and then restrict to an irreducible closed subscheme; this includes reduction modulo primes. If specialization results in a separable polynomial of the same degree over the new function field, then the Galois group of the specialization is a subgroup of the original Galois group, so it suffices to show that the specialized Galois group is not contained in $A_n$.
For more techniques for computing Galois groups, I recommend the book Topics in Galois theory by Jean-Pierre Serre.