Assume a collision-resistant hash function $h$ which compresses sequences of length $2n$ to length $n$. It must be examined whether the hash function which compresses seqs. length $4n$ to $n$:
$$ h_1(x_1 || x_2 || x_3 || x_4) = h((x_1 \oplus h(x_2 || x_2)) || (h(x_3 || x_3) \oplus x_4)) $$
is collision-resistant or not. ($\oplus$ is XOR, $||$ is concatenation and $|x_i| = n$).
Assume we can easily find two distinct sequences $(y_1, y_2) = (x_1 || x_2 || x_3 || x_4, x_5 || x_6 || x_7 || x_8)$ such that $h_1(y_1) = h_1(y_2)$.
Then, we can easily find $(y_1, y_2)$ such that:
$$ h(g(y_1)) = h(g(y_2)) $$
where $g(y) = g(x_1||x_2||x_3||x_4) = (x_1 \oplus h(x_2 || x_2)) || (h(x_3 || x_3) \oplus x_4)$.
But then, if we can easily find such $(y_1,y_2)$, we can also easily find $(g(y_1), g(y_2))$ and thus $h$ is not collision-resistant.
In case the above argument isn't valid, another idea would be to state that $g(y_1), g(y_2)$ cannot be distinct (with non-negligible probability) since $h$ is collision-resistant and continue with solving $g(y_1) = g(y_2)$ etc until we get $y_1 = y_2$.
It'd be very helpful if someone pointed out to me which of the two (if any) is the correct proof.