The birthday "paradox" asserts that if $$$n$$$ people choose one element each from $$$k$$$ possible elements independently, there will be a collision (two people choosing the same element) with high probability provided that $$$n^2 » k$$$. This is due to the fact that there are $$$n*(n-1)/2$$$ different couples that may collide. Thus if you have a group of 23 people there will be at least two of them with the same birthday with a probability higher than 0.5. In this case $$$k = 356$$$ and $$$n = 23$$$.

We are goint to exploit this fact to solve the problem that can be formulated as follows: You have 4 sequences of random n-bit numbers $$$X_1$$$, $$$X_2$$$, $$$X_3$$$, $$$X_4$$$. You want to find four elements $$$x_1\in X_1$$$, $$$x_2 \in X_2$$$, $$$x_3\in X_3$$$, and $$$x_4 \in X_4$$$ such that $$$x_1 \oplus x_2 \oplus x_3 \oplus x_4 = 0$$$.

We proceed as follows. Consider a list of about $$$2^{n/3}$$$ numbers from $$$X_1$$$ and $$$2^{n/3}$$$ number from $$$X_2$$$. There are about $$$2^{2n/3}$$$ couples so we may expect that there are $$$2^{n/3}$$$ couples such that that their first $$$n/3$$$ bits match ($$$k = 2^{n/3}$$$). So we find a list of $$$2^{n/3}$$$ couples $$$x_i, x_j$$$ such that $$$x_i\oplus x_j$$$ is zero in its first $$$n/3$$$ bits and $$$x_i\in X_1$$$ and $$$x_j \in X_2$$$.

We do the same with $$$X_3$$$ and $$$X_4$$$ and we find $$$2^{n/3}$$$ couples such that their first $$$n/3$$$ bits match.

Finally again by the birthday paradox we may expect that there are a couple of the first group and a couple from the second group such that their corresponding xor matches on the last $$$2n/3$$$ bits.