There's already been some good answers to this. However, this is something that I have also thought about recently, because I happen to have come across several meagre sets of full Lebesgue measure in some of my answers to other questions. In fact, in my experience on MO, meagre sets with full Lebesgue measure actually seems to be more the rule than the exception. So, I'll add these to the list.
In this math.SE question and this MO questionthis MO question, David Speyer was trying to find the set of θ such that $\sum_{n=1}^\infty \sin(n^r\theta)/n$ converges (r > 1 an integer). He was concerned about the θ = 1 case but, from my answer on MO and David's answer on math.SE it can be seen that it converges for almost every θ but, at the same time, it only converges for θ in a meagre set.
Along a similar line, this MO questionthis MO question was asking for which θ the asymptotic bound $\sum_{n=1}^N{\rm sign}(\sin(n\pi\theta))=O(N^x)$ holds. For 1/2 < x < 1 my answer shows that it holds for almost every θ but, at the same time, it only holds for θ in a meagre set.
This questionThis question asks if there are 2x2 matrices C such that Tr(Cn) is dense in the reals as n runs through the positive integers. Bjorn Poonen shows that the answer is yes. In fact, his proof is easily modified to show that Tr(Cn) fails to be dense only on a meagre set. However, my answer shows that |Tr(Cn)| is either bounded or tends to infinity (so, not dense) for almost every C.
The above examples really just come down to the following point. The set of real numbers with finite irrationality measure (i.e., non-Liouville numbers) is meagre. However, almost every real number has irrationality measure 2.
Along similar lines, the set of normal numbers is meagre and has full Lebesgue measure (see also, Andreas' answerAndreas' answer). The set of real numbers whose continued fraction quotients have geometric mean converging to Khinchin's constant is meagre with full Lebesgue measure. The set of real numbers whose continued fraction quotients occur according to the Gauss-Kuzmin distribution is meagre with full Lebesgue measure. And so on...