How to use generalized pigeonhole principle to be sure that at least one of the integers picked is even? 
Q. How many integers from $0$ through $60$ must you pick in
  order to be sure of getting at least one that is odd? at least
  one that is even?

Here is my verbal solution 'without using' pigeonhole principle.

A. The list of integers $0,1,2...,60$ has $31$ even integers $30$ odd integers. So in order to be sure that at least one of the picked integers is odd, we must pick $32$ integers. This is because it may happen that first $31$ integers we pick turn out to be even. By similar argument, we find that if we pick $31$ integers from the list, then we are sure that at least one amongst them is even.

Now I want to write this solution in terms of a function, pigeons and pigeonholes.
Unfortunately the list of given integers contains total $61$ integers which is odd. Thus I can't take pigeonholes to be $\{0,1\}, \{2,3\}, \{4,5\},...,\{58,59\},\{60\}.$ The problem is that last element is a singleton set. I tried to use these pigeonholes as follows:

Let $X$ denote the set of pigeons that are picked and $Y=\{A_1=\{0,1\},A_2= \{2,3\},A_3= \{4,5\},...,A_{30}=\{58,59\},A_{31}=\{60\}\}$ denote the set of pigeonholes. Let $f : X \to Y$ be a function such that $f(x)=A_i$ whenever $x \in A_i.$
To be sure that at least one integer is odd : Suppose that we have picked $n$ number of pigeons. Then we are through when we find the least value of $n$ such that $\lceil \frac n{31} \rceil \ge 2$ (By generalized pigeonhole principle). Observe that $n=32$ fits the bill. Also $A_{31}$ is not the pigeonhole containing at least $2$ pigeons since the set $\{60\}$ has only one element. Thus one amongst the other pigeonholes must work out and we are sure that we have found at least one odd integer.
However, this argument fails when we want to be sure that at least one integer is even.

How should I think about pigeons and pigeonholes here?
EDIT: Should I consider different pigeonholes for the case of picking at least one even integer? Specifically I would just change $A_{30}=\{58,59\}$ to $A_{30}=\{58,59,60\}$ and rule out $A_{31}$ for this case. Then the least value of $n$ such that $\lceil \frac n{30} \rceil \ge 2$ is $n=31.$ Here even if $A_{30}$ turns out to be the pigeonhole with at least $2$ pigeons we are sure that one of the pigeons is bound to be even since this pigeonhole has only one odd number viz $59$ in it.
 A: Actually, your initial reasoning is a perfectly good instance of 'reasoning by pigeonholing':  there are at most 31 'even' holes for pigeons to go in, so with 32 pigeons you're bound to get an odd number. 
That's it!
Your second method is far more complicated than it has to be. Yes, you can make it work by making the holes $\{ 0,1 \}$, $\{2,3 \}$, etc. but also by using $\{0,3 \}$, $\{ 1,2 \}$, etc. In fact, to get as many holes as even numbers, you could even use $\{ 0,37,39 \}$, $\{2, 13 \}$, $\{ 4 \}$, $\{ 6, 19,23,29,59 \}$, etc.  In other words, adding the odd numbers to the even numbers when all that really counts is how many even numbers there are is completely extraneous.
Now: I understand you tried to set it up in such a way that you can try to answer both the question about the odd and the even numbers at once ... which seemed to work fine ... until you got to the $60$ 'by-itself-hole' ... and now you get into trouble: Using $\{ 60 \}$ as a hole means it can contain exactly one 'even' pigeon, but now it cannot contain any 'odd' pigeon, and so it can't be used to do the reasoning regarding odd numbers, and using $\{ 58,59,60 \}$ means two 'even' pigeons can go into that hole, and so now it cannot be used to reason about the even numbers. 
So really the best thing is to answer the question about the even numbers separately from the question about the odd numbers: with $31$ even numbers you need to pick $32$ pigeons to get an odd pigeon, and with $30$ odd numbers you need $31$ pigeons to get an even pigeon.
A: I'll attempt to answer the question in the sense you mean to ask it.
As noted already, your first argument actually is what many people
(myself included) already would consider an argument by the pigeonhole principle.
The second argument adds a lot more structure on top of this, namely the partition of the set $\{0,\ldots,60\}$ into $31$ subsets (to prove that you can choose $31$ even integers but not $32$) or $30$ subsets (to prove that you can choose $30$ odd integers but not $31$).
The extra structure adds information to any particular selection of numbers that is not really relevant to the thing to be proved;
namely, if you have already chosen $31$ even integers, the partition tells you not only that the $32$nd number will be assigned to an already-occupied pigeonhole; once you actually choose a $32$nd number,
the partition tells you specifically which odd number is already in the pigeonhole where you are trying to put the new number.
I don't see anything particularly wrong with this;
I'm sure there are plenty of proofs where we build some structure that does a little more than the thing to be proved, because it's a convenient way to get at least what we needed.
And you may have good reasons to want a partition
(for example, if the principle has been defined in terms of partitions).
I would say that the partition for "$32$ numbers must include one odd number" has just a few essential properties:


*

*It consists of exactly $31$ subsets.

*Each subset contains exactly one even number.


Everything else is arbitrary, so you can arrange it in whichever way you want. A perfectly good partition is to have $30$ singleton subsets
containing one even number each, and
everything else in the last subset:
$$\{0\},\{2\},\{4\},\ldots,\{58\},\{60, 1,3,5,\ldots,59\}.$$
So now we have the (unnecessary, but harmless) extra information that if we selected two numbers for the same pigeonhole, they're both in the pigeonhole that was assigned to the number $60.$
And those double-occupancy numbers are either two odd numbers (so you have chosen at least one!) or $60$ and an odd number.
Since there's no problem putting $31$ numbers in one subset of this partition, I would be equally willing to put three numbers in one of the $30$  subsets, namely $A_{30}=\{58,59,60\}$, when making a partition for a proof that at least one number must be even if you choose $31$ numbers.
