Suppose we have a set of 25 distinct positive real numbers. Show that one can always choose 2 of them such that neither their sum, nor their difference, is equal to one of the other 23 numbers in the set.
It seems like a pigeonhole problem and I guess I need to use proof by contradiction however I can't figure out such construction to use the principle.
First, sort these $25$ distinct positive real numbers into increasing order and name them $r_i, \; 1 \le i \le 25$.
As you suggested, a proof by contradiction can be used here. In particular, assume there exists a set of these $25$ numbers where you can't choose $2$ of them where neither their sum nor their difference is equal to one of the other $23$ numbers in the set. In particular, this means that no matter which $2$ you choose, either their sum or their difference must be one of the other $23$ numbers in the set.
Consider choosing $r_{25}$, i.e., the largest value, as one of the $2$ values. Since it's the largest, adding any of the other $24$ positive values to it would make it larger than $r_{25}$. Thus, it must be the difference between $r_{25}$ and the other value which is another value in the set. Consider the set of differences $r_{25} - r_i$ for $i = 24, 23, \ldots, 1$. Since $r_i$ is decreasing as $i$ decreases, the difference must be increasing, and each value is unique, with none of them being $r_{25}$ as $r_i \gt 0$. Thus, you have $24$ increasing unique values which match among the $24$ other $r_i$ values. This can only happen if each value is used exactly once in order, i.e., you have
$$r_{25} - r_i = r_{25-i}, \; 1 \le i \le 24 \tag{1}\label{eq1A}$$
In particular, note you have
$$r_{25} - r_{24} = r_1 \implies r_{25} = r_{24} + r_1 \tag{2}\label{eq2A}$$
Next, choose $r_{24}$ as one of the values, and consider each $r_i, \; 2 \le i \le 23$. Since each of these $r_i \gt r_1$, this means from \eqref{eq2A}, their sum would be greater than $r_{25}$ and, thus, not part of the set. As such, each it's each difference that must be a value in the set.
From \eqref{eq1A}, you also have
$$r_{25} - r_{23} = r_2 \tag{3}\label{eq3A}$$
Now, \eqref{eq3A} minus \eqref{eq2A} gives
$$r_{24} - r_{23} = r_2 - r_1 \tag{4}\label{eq4A}$$
Since $r_2 - r_1 \lt r_2$, for it to be part of the set means that
$$r_2 - r_1 = r_1 \implies r_2 = 2r_1 \tag{5}\label{eq5A}$$
Also from \eqref{eq1A}, you have
$$r_{25} - r_{22} = r_3 \tag{6}\label{eq6A}$$
Next, \eqref{eq6A} minus \eqref{eq2A} gives
$$r_{24} - r_{22} = r_3 - r_1 \tag{7}\label{eq7A}$$
Since $r_3 - r_1 \lt r_3$, it requires $r_3 - r_1 = r_2$ or $r_3 - r_1 = r_1$. It can't be the second one since that leads to $r_3 = 2r_1 = r_2$, but $r_3 \gt r_2$. Thus, it requires that
$$r_3 - r_1 = r_2 = 2r_1 \implies r_3 = 3r_1 \tag{8}\label{eq8A}$$
You can repeat this process for each remaining $r_i$ subtracted from $r_{24}$ from $r_22$ down to $r_1$ (e.g., just manually or by using induction) to get
$$r_i = ir_1, \; 1 \le i \le 23 \tag{9}\label{eq9A}$$
Next, from $r_2 - r_1 = r_1$ in \eqref{eq4A}, this gives that
$$r_{24} - r_{23} = r_1 \implies r_{24} = 23r_1 + r_1 = 24r_{1} \tag{10}\label{eq10A}$$
Finally, from \eqref{eq2A}, you have
$$r_{25} = 24r_1 + r_1 = 25r_1 \tag{11}\label{eq11A}$$
This means $r_i = ir_1$ for $1 \le i \le 25$. However, consider now choosing $r_9 = 9r_1$ and $r_{18} = 18r_1$ (or, equivalently, $r_{10}$ & $r_{20}$, $r_{11}$ & $r_{22}$, or $r_{12}$ & $r_{24}$). Their difference is $9r_1$ and their sum is $27r_1$, with neither value being one of the other $23$ values. In particular, the difference is one of the $2$ values chosen and the sum not even in the set.
Thus, this contradicts the original assumption that you can't choose any such $2$ values among the set, proving this assumption must be incorrect. As such, using proof by contradiction, you have that among any $25$ distinct, real positive numbers, you can always choose $2$ of them such that neither their sum nor neither difference is equal to one among the other $23$ numbers in the set.
Update: Note the above proof doesn't depend on their being $25$ elements in the set. In fact, if $n$ is the number of elements, then the above proof by contradiction would end up requiring $r_i = ir_1, \; 1 \le i \le n$ for any $n \ge 1$. Also, any $n \ge 4$ works as you can get a contradiction by choosing, if $n$ is even the elements $r_{n/2}$ and $r_{n}$ else if $n$ is odd the elements $r_{(n-1)/2}$ and $r_{n-1}$. Some possible reasons why $25$ was chosen is that they didn't want a value too small (e.g., just $4$ or $5$) that the students could figure it out using just several equations, there's another proof method which requires the value to be at least (or even exactly) $25$, and that $25$ was just more or less picked randomly.