The Wikipedia article explains exactly what it means. Here is an elaboration.
Define the complexity of a number $n$ to be the smallest number of states for a Turing machine that returns $n$ when run with $0$ as an input. Certainly every number has a complexity. Now let $T$ be any effective formal theory whose language includes statements of the form "$n$ has complexity greater than $k$" for each $n$ and $k$. Some of these sentences may be provable in the theory, and some may not. Say that the theory is complexity sound if each statement "$n$ has complexity greater than $k$" that is provable in the theory is correct. Both PA and ZFC are effective, complexity sound theories.
Theorem. For any effective, complexity sound theory $T$, there is some $k$ such that the theory does not prove that any number $n$ has complexity greater than to $k$.
Proof by contradiction. Assume $T$ is complexity sound and proves statements of the form "$n$ has complexity greater than $k$" for arbitrarily large $k$. Consider a program $P$ that does the following when run with $0$ as an input. First, $P$ computes its own source code (like a Quine; formally, the proof uses Kleene's recursion theorem). Then $P$ counts the number of states $s$ in this source code. Next $P$ enumerates all statements of the form "$n$ has complexity greater than $k$" that are provable from $T$, until it finds one with $k > s$. It will find one by assumption. Finally, $P$ returns $n$. Because $T$ is an effective theory, $P$ can do all this computably. But then the number $n$ returned is by $P$, which has $s$ states, but $T$ proves "$n$ has complexity greater than $k$" for some $k > s$, so $T$ is not complexity sound, which is a contradiction. That completes the proof.
Thus there is a bound on the largest $k$ such that $T$ proves "$n$ has complexity greater than $k$" for any $n$. In fact, we can actually write down $P$ explicitly, and thus $k$ is no more than the number of states of the version of $P$ we write down. For any reasonable theory, like PA or ZFC, this $k$ will be huge but not anywhere near $10\uparrow\uparrow 10$.
If a sufficiently strong theory cannot prove any $n$ has complexity larger than $k$, then it also cannot prove any statement of the form "$m$ is an upper bound on $\Sigma(k)$", because if it could prove that it would also prove "$m+1$ has complexity greater than $k$". So in particular neither PA nor ZFC can prove an upper bound on $\Sigma(k)$ when $k$ is sufficiently large.
Just as the Wikipedia article says, the theorem above is just a recasting of Chaitin's incompleteness theorem in which the length of the program is replaced with the number of states.