Consider the sequence of initial sums $S(0),S(1),\ldots,S(55)$
$$
S(n)=\sum_{i=1}^na_i.
$$
We have
$$S(0)=0<S(1)<S(2)<\cdots<S(55)<95,$$
a sequence of 56 distinct integers in the interval $[0,94]$ of $95$ possibilities.
Consider also the sequence of numbers $T(i)=S(i)-15$. There are 56 of those as well, all integers in the range $[-15,79]$. Because they are all distinct, at least $56-15=41$ of them are non-negative.
Hints:
- Can you show that the sets $\{S(i)\mid 0\le i<56\}$ and $\{T(i)\mid 0\le i<56\}$ must have a non-empty intersection?
- Do you see how the claim follows from this?
Spoiler #1
Between them the two sets have 97 positive integers in the range $[0,94]$ so the pigeonhole principle tells us that there is some overlap.
Spoiler #2
If $T(\ell)=S(\ell)-15=S(k)$ then $$ \sum_{i=k+1}^\ell a_i=S(\ell)-S(k)=15.$$
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.
Best Answer
For each $1\leq i\leq 55$ define $x_i=a_1+a_2+...+a_i$. Since the integers are positive we know that $1\leq x_1<x_2<...<x_{55}\leq 94$. Now look at the following set:
$A=\{x_1,...,x_{55},x_1+15,...,x_{55}+15\}$
The number of elements in $A$ is at most $94+15=109$, because every element is an integer in $\{1,2,...,109\}$. But from the way I defined the set it looks like there are $110$ elements. From the pigeonhole principle we get that there must be a duplicate. Since $x_1,...,x_{55}$ are all different elements, and so are $x_1+15,...,x_{55}+15$ we have to conclude that there are some $i,j$ for which $x_i=x_j+15$. And that gives us $15=x_i-x_j=a_{j+1}+...a_{j+2}+...+a_i$.