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Suppose we have a basic statement that could easily be proved directly. For example,

Prove that if n is an even integer, then 7n-9 is an odd integer.

This could be proven by saying that n = 2k for some k and stating that 7n-9 = 7(2k)-9 = 14k-9 = 2(7k-5)+1, which is odd.

However, how would you prove something like this by contrapositive? Is it even possible?

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    To prove it by contrapositive you would need to prove that if $7n-9$ is not odd then $n$ is not even. Ie if $7n-9$ is even then $n$ is odd. – 2017-02-19
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    @lordoftheshadows Caution: "not an odd integer" does not mean "an even integer". It might not be an integer at all. So you need to know the fact that if $n$ is an integer, $7n-9$ is an integer. – 2017-02-19
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    @RobertIsrael good point. Although I would assume that the closure of the integers under addition is assumed/proved in your average discrete course. – 2017-02-19
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    Anything of the form $p\rightarrow q$ is equivalent to its contrapositive: $\lnot q \rightarrow \lnot p$. Proving $p\to q$ is equivalent to proving $\lnot q\to \lnot p$. Assuming $n \in \mathbb Z$, then we have to equivalent statements: "If n is an even integer, then $7n-9$ is odd", and "If $7n -9$ is even, then n is odd. – 2017-02-20
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    The negation of n is odd is n is an even integer or n is not even an integer which is odder. – 2017-02-20

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Of course, although unnecessary convoluted.

Let n be an integer.

Let a=7n-9 be not an odd integer, prove that n is not even.

Since a is obviously an integer, and not odd, it must be even, thus a+9 =7n is odd (odd+even =odd). Since even x odd = even, and the result of 7n is odd, n must be odd that is not even.

Now, consider if a = 7n-9 is non-odd but n is even - according to just given proof, it's not possible. Thus, if n is indeed even, 7n-9 must be odd, q.e.d.