Is the contrapositive of a true statement always provable? I had a very difficult time proving the contrapositive of a simple elementary number theory statement, and the more I think about, the more I believe it might not be provable that way using any proof techniques (that we accept as valid). I know there are true statements that aren't provable, but I'm not sure how much it relates to this. I have heard of something about models, or say, a model of Number Theory: NT. Are there generalizations of types of true statements we can prove in number theory in certain ways but not in others?
 A: As others have noted in the comments, using the adjective true opens a can of worms: truth has a technical definition in semantics, and it almost certainly clashes with the question you want to ask. In particular, true things need not be provable, period.
Instead, let me state a proof-theoretic question that (I think) cuts close to the one you intend to ask:

I have managed to prove an implication $A \rightarrow B$, but I could do it only by taking the contrapositive $\neg B \rightarrow \neg A$, and proving that. In such a situation, can I always find a a more direct proof of $A \rightarrow B$ that does not involve taking contrapositives?

Based on your question, your gut feeling tells you that this need not be the case, and your gut feeling is correct. Sometimes, every proof of a statement requires you to take contrapositives.
Before I explain why, I have to explain some things regarding proof-theoretic terminology. To investigate proofs rigorously, we first have to define what we mean by proof. There is a large variety of such definitions, called proof systems (deductive systems).
Some proof systems define obscure or extremely constrained forms of reasoning, such as pedagogical reasoning, where you have to give explicit examples before introducing abstract concepts; or ethical reasoning, where you have to distinguish between permissible and forbidden statements. Different proof systems may prove different theorems. But even two proof systems which prove the same theorems can differ substantially from each other: a theorem can have multiple proofs in one system, yet only one proof in some other system. Proof theorists may whip up new deductive systems on demand, similarly to the way a group theorist would construct new groups to illustrate technical situations or to provide (counter)examples to mathematical conjectures and questions.
With these differences in mind, it should be clear that we'll have to fix a proof system up front to say anything sensible. From here on, I will focus on one specific proof system: the natural deduction of Gentzen and Prawitz. Natural deduction provides a rigorous substrate to the kind of proof you're likely to be interested in: mathematical proof of the kind that would be accepted in an ordinary mathematics textbook or journal article.
II. Writing a thorough introduction to natural deduction, let alone one that fits in a Math.SE answer, is a considerable challenge, so much so that I won't even try. If you can read and write mathematical proofs, you are equipped to understand natural deduction. You might be able to pick up how it works based purely on what's written below, but if you wish to learn all the rules and understand it more thoroughly, there are many tutorials, YouTube videos, and myriads of textbooks on the subject. I personally recommend the first three chapters of Jan von Plato's Elements of Logical Reasoning.
Natural deduction has many inference rules, all of them familiar to the working mathematician. These inference rules tell you how to construct new proofs from existing proofs. Each connective (conjunction $\wedge$, disjunction $\vee$, implication $\rightarrow$, negation $\neg$) and quantifier (universal $\forall$, existential $\exists$) comes equipped with one or more so-called introduction rules and one elimination rule. Examples:

*

*If you have a proof of $A$ and you have a proof of $A \rightarrow B$, then you have managed to prove that $B$ holds (implication elimination).

*If you have a proof that $A \wedge B$ ($A$ and $B$) holds, then you have managed to prove that $B$ holds (conjunction elimination, right).

*If you have a proof that $A \wedge B$ holds, then you have managed to prove that $A$ holds (conjunction elimination, left).

*If you have a proof that $A$ holds, and you have a proof that $B$ holds, then you have managed to prove that $A \wedge B$ holds (conjunction introduction).

*If you have a proof that starts with "Suppose $A$..." and ends with "... therefore $B$", then you have managed to prove the conditional statement $A \rightarrow B$ (implication introduction).

*If you have a proof that starts with "Suppose $A$..." and ends with "which is a contradiction", then you have managed to prove the negation $\neg A$ (negation introduction).

*and so on...

Beyond these ordinary rules, there are three special rules as well:

*

*If you have a proof that $\neg\neg A$ holds, then you have managed to prove that $A$ holds (double negation elimination).

*You always have a proof of $A \vee \neg A$ (law of excluded middle).

*If you have a proof that $\neg B \rightarrow \neg A$, then you have managed to prove that $A \rightarrow B$ (proof by contrapositive).

If you add any one of these rules to the ordinary rules of natural deduction, you can prove the other two. For example, if you add the law of excluded middle as a rule of inference, you can prove every instance of double negation elimination. Most importantly for us, you can use proof by contrapositive to prove every instance of double negation elimination. We can use this auxiliary proof:

*

*Suppose that $\neg A$ holds.

*Suppose that $\neg\neg A$ holds.

*From 1 and 2 we have a contradiction.

*From 2-3 we have $\neg\neg\neg A$ by negation introduction.

*From 1-4 we have $\neg A \rightarrow \neg\neg\neg A$ by implication introduction.

*From 5 we have $\neg\neg A \rightarrow A$ by contrapositive.

Now, imagine that we have a proof of $\neg\neg A$. Our auxiliary proof gives us $\neg\neg A \rightarrow A$, so implication elimination gives us a proof of $A$. So, as I claimed, you can use the proof by contrapositive inference rule, along with the ordinary rules, to prove every instance of double negation elimination.
The natural deduction proof system that has all the ordinary introduction and elimination rules, but none of the three special rules, is called Intuitionistic or Constructive Natural Deduction. We sometimes call the system that includes one of the special rules as well (most commonly double negation elimination, for technical reasons) Classical Natural Deduction. It is a well-known fact that $\neg\neg A \rightarrow A$ is not provable in Intuitionistic Natural Deduction. So in a very real sense, $\neg\neg A \rightarrow A$ is not provable without taking contrapositives, or one of the other two rules of inference equivalent to it.
III. If you put the axioms of elementary number theory into an Intuitionistic Natural Deduction proof system, you end up with a mathematical theory called Heyting Arithmetic. Keep in mind: just because a general logical principle is not provable without taking contrapositives, it does not follow that no instance is provable without taking contrapositives! E.g. Heyting arithmetic proves $\neg\neg t = 0 \rightarrow t = 0$ for any number $t$.
However, Heyting Arithmetic differs from usual (Peano) elementary arithmetic in a number of ways. In particular, Heyting Arithmetic does not prove the following:

For every polynomial $P$ with integer coefficients, there are integers $n$ such that for all $x$, $|P(n)| \leq |P(x)|$. [2]

This provides an answer to one of the weaker formulations of your question: every proof of the theorem above requires you to take contrapositives (or use an equivalent principle) somewhere in its proof.
I won't go into details, but a proof-theoretic result of De Jongh [1] allows us to answer a stronger formulation: we can also obtain explicit number-theoretic statements $A,B$ such that Heyting arithmetic proves the implication $\neg B \rightarrow \neg A$, but not the implication $A \rightarrow B$.
You could ask an even stronger question: find explicit number-theoretic statements $A,B$ such that every (sufficiently normalized) proof of $A \rightarrow B$ contains a proof of $\neg B \rightarrow \neg A$ itself as a subproof. Good luck with that: it sounds very very difficult :)

[1] De Jongh, D.H.J.: The maximality of the intuitionistic predicate calculus with respect to Heyting's arithmetic, tech. rep., Meeting of the Association for Symbolic Logic, Manchester UK, 1969
[2] Friedman, H.: Classical/Constructive Arithmetic, FOM mailing list, 18 Mar 2006
