Consider the set $$A = \{1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34\}$$ which consists of all natural numbers of the form $3k + 1$ which are less or equal to $34$. What is the smallest number $n$ that has the following property: if we choose $n$ numbers from $A$ randomly, then there will always be a pair of numbers (in these $n$ numbers) such that their sum is 35.
How is this question approached using the Pigeonhole Principle?
You've written the elements in increasing order. Number them in this order, so that $1$ is the first element, $4$ is the second, and so on, up to $34$ is the twelfth. Note that, for all $k=1,2,\ldots,12,$ we have $$k\text{-th element}+(12-k)\text{-th element}=35.$$ This gives six pairs adding to $35.$ So if you pick $7$ elements, you must have picked at least one pair together (by the Pigeonhole Principle); but if you pick just $6$ elements, you might not have picked a pair (for example, your list of six elements could be $\{1,4,7,10,13,16\}$). Hence $n=7$ is the smallest.
See that $$35=1+34=4+31=7+28=...=16+19$$
And we have $6$ pairs of that.
So, by Pingeonhole, the minimum value of $n$ is $7$. Because if you choose just $6$ numbers you can have a bad luck and find the set $\{1,4,7,10,13,16\}$, but if you pick up just one more you surely have one of those pairs.