I do not know if you are interested in hearing more detail of my comment on your question, but here is more detail anyway.
If P=NP, every problem in NP can be solved in polynomial time and therefore in pseudo-polynomial time, which means that no problem satisfies your requirement, as Magnus noted in his answer. So assume P≠NP in the rest of this answer.
Because P≠NP, there exists a language L∈NP∖P which is not NP-complete (Ladner’s theorem). Consider the following problem:
Direct product of Partition and L
Instance: m positive integers a1, …, am and k integers b1, …, bk∈{0,1}.
Question: Do both of the following hold?
(1) The m integers a1, …, am form a yes-instance of the Partition problem.
(2) The k-bit string b1…bk belongs to L.
Following the paper by Garey and Johnson, define the Length function as m+⌈log maxi ai⌉+k and the Max function as maxi ai.
It is a routine to check (i) that it is NP-complete in the weak sense, (ii) that it does not have a pseudo-polynomial-time algorithm, and (iii) that it is not NP-complete in the strong sense.
(Hints: (i) Membership to NP follows from the fact that both the Partition problem and L are in NP. For NP-hardness, reduce Partition to this problem. (ii) Construct a pseudo-polynomial transformation from L to this problem. (iii) Construct a pseudo-polynomial transformation from this problem to L by using the fact that Partition has a pseudo-polynomial-time algorithm.)
There is nothing special about the Partition problem in this construction: you can use your favorite weakly NP-complete problem with a pseudo-polynomial-time algorithm.