Let's translate the problem into the language of graphs.

Given a bipartite graph $G$ with parts $A$ and $B$ having the same number of vertices: $|V(A)| = |V(B)| = k$. The degree of each vertex in $B$ is $n$. For any two vertices $u, v$ of the part $A$, there are exactly two vertices in $B$ that are adjacent to both vertices $u$ and $v$.

a) Prove that the degree of any vertex in $A$ is also equal to $n$.

b) Prove that there is a simple path in the graph that contains at least $\lfloor \frac{(n+1)^2}{4} \rfloor$ vertices in each part.

First, let's prove that $k = \binom{n}{2} + 1$ and that the degree of each vertex in $A$ is equal to $n$.

Let's count the number of pairs ($\{a, a'\}, b$) where $a \neq a'$, $a, a' \in V(A)$, $b \in V(B)$ and $b$ is adjacent to $a$ and $a'$. On the one hand, we have $|V(B)|$ ways to select $b$ and each such vertex gives $\binom{n}{2}$ pairs $a, a'$ to which it is adjacent. On the other hand, there are $\binom{|V(A)|}{2}$ ways to select distinct $a$ and $a'$, and further, for each pair, there are exactly two ways to select $b$, according to the second property of our graph. Thus, $\binom{n}{2}k = 2\binom{k}{2}$, which implies that $k = \binom{n}{2} + 1$.

Let's take an arbitrary vertex $a$ from $A$ and let it have degree $d$. Let us count in two ways the number of pairs $(a', b)$ where $a' \in V(A)$, $b \in V(B)$, $a' \neq a$ and vertex $b$ is adjacent to both vertices $a$ and $a'$. On the one hand, we have $|V(A)| - 1$ options to choose a vertex $a'$ and for each such vertex we have exactly two choices of a vertex $b$. On the other hand, each of the $d$ neighbors of a vertex $a$ has degree exactly $n$, which gives $(n-1)d$ pairs in question (the neighbors of a given vertex from $V(B)$ must be different from $a$, so there are $n-1$ of them). Thus $(n-1)d = 2(k-1) = 2\binom{n}{2}$, which means that $d = n$.

Now let's move on to finding a path. Since the graph is regular, then by Hall's theorem there is a perfect matching in it – a set of $k$ pairwise non-adjacent edges. Let $I$ be such a matching. Let us take the longest such path $P = b_1a_1, \dots, b_ra_r$, where $b_1 \in V(B)$ and $a_r \in V(A)$, with the following property:

if a vertex $x$ is in $P$, then the vertex $y$ adjacent to $x$ in the matching $I$ is also in $P$.

(Note that this condition does not require that edges from $I$ also be in our simple path.) Then all neighbors of the ends of $P$ lie in $P$. In fact, if one of the ends of $P$ has a neighbor outside $P$, then the neighbor of this vertex in $I$ is also not in our path and then we can extend our path by two vertices so that it starts and ends at different shares.

Let now $b_i$ ($1 \le i \le r$) be one of the neighbors of a vertex $a_k$ in path $P$. Then the path
$$b_1a_1 \dots b_ia_r b_r a_{r-1} \dots a_i$$
also has the above property. In particular, all neighbors of the vertex $a_i \in V(A)$ also lie in $P$. Since $a_r$ has $n$ neighbors, we have a minimum of $n$ vertices $a_{i_1}, \dots, a_{i_n}$ from $A$ each of which has neighbors only in the path $P$. Since vertices $a_{i_1}$ and $a_{i_2}$ have exactly two neighbors in common, they have a total of $n + (n-2)$ neighbors in $P$. Next, consider the vertex $a_{i_3}$ – it has two common neighbors with each of the previous vertices, that is, a maximum of 4 common neighbors have already been counted. This gives $n-4$ new neighbors different from the previous ones. Reasoning in this way, we get
$$r \ge n + (n-2) + (n-4) + \dots = \left\lfloor \frac{(n+1)^2}{4} \right\rfloor,$$
which is what was required.