Let T be a first order set theory formalized in the language L(ZF) of ZF, which has "membership" and "=" as its only atomic predicates. For each positive integer n, let P(n) be the sentence which expresses that "there exists a set having exactly n elements". P(n) can be formalized in L(ZF) and is an axiom of T for each positive integer n. Note that L(ZF) does not need to contain any terms that are constants for this to be possible. Now let Q be the sentence stating that "there exists a finite set (in Tarski's sense of finite) which can be mapped onto every nonempty set". Q is also an axiom of T that can be formalized in L(ZF) and should be consistent with every finite collection of sentences of the form P(n). However with infinitely many axioms it would seem appropriate to call T at least a winconsistent theory. But could T still be consistent? The answer is not clear to me since it would depend upon what other axioms T has. T must have some other axioms (such as the axiom of pairing) in order to define the mapping of one set onto another as the sentence Q describes. We must be careful that these additional axioms are not inconsistent with Q. We should probably not want a power set axiom, for example. But, with a proper choice of the new axioms could we end up with a consistent T? If so, T would be an example of a consistent but winconsistent formalized theory whose language need contain no constants and no formulae of the form N(x) intended to be interpreted as "x is a positive integer".

$\begingroup$ Tarski finiteness is somewhat of an ambiguous term. One definition is equivalent to finiteness as we think about it, and the other one requires the use of the axiom of choice in order to be equivalent to finiteness. $\endgroup$ – Asaf Karagila Mar 11 '13 at 21:17
If we replace your axiom $Q$ by the strongerseeming assertion $Q^+$ that asserts: "there is a largest set, which contains all other sets as subsets, but which is Tarski finite", then the corresponding theory $T$, asserting every $P(n)$ and also $Q^+$, is consistent, but not $\omega$consistent.
To see this, simply consider the ultrapower model $M=\Pi \langle V_m,{\in}\rangle/U$, where $U$ is a nonprincipal ultrafilter on $\mathbb{N}$ and $V_m$ consists of the sets of rank less than $m$ in the von Nuemann hierarchy, defined by iterating the power set via $V_0=\emptyset$ and $V_{m+1}=P(V_m)$. Each statement $P(n)$ is true in all but finitely many $V_m$, and hence true in $M$. Similarly, each $V_{m+1}$ satisfies $Q^+$, since the object $V_m$ is largest in $V_{m+1}$ and Tarski finite there, and so $Q^+$ also is true in $M$.
Alternatively, one could consider $V_N$ for nonstandard $N$ inside a nonstandard model of finite set theory $\text{ZFC}^{\neg\infty}$.
These models furthermore satisfy some nice axioms, such as extensionality, union, foundation, axiom of choice, collection, replacement, separation, pairing restricted to sets of nonmaximal rank, power set restricted to sets of nonmaximal rank and induction on the von Neumann natural numbers. And so one can place all these axioms also into $T$.
This theory, however, is clearly not $\omega$consistent.

$\begingroup$ Thanks, Joel, for a very complete answer. Your axiom Q+ accomplishes everything that axiom Q does, while avoiding the problem of some additional axioms being possibly needed to define mappings. Nothing new would seem to be needed to express that a set is finite in the sense of Tarski. So, if the only axioms of T are axiom Q+ and an infinite collection of sentences of the form P(n), T may be one of the simplest possible examples of a formalized theory that is both consistent and winconsistent. $\endgroup$ – Garabed Gulbenkian Mar 13 '13 at 18:51