Taiwan IMO Selection Camp

Consider a way of selecting $s$ edges satisfying the condition, and color these $s$ edges red, and the other $n-s$ edges blue. If the red edges are not connected, then adding an edge connecting different connected components must not produce a new cycle, which contradicts the maximality of $s$, so these red edges are connected and acyclic, hence these $s$ edges form a tree. Now consider the blue edges $b_1, b_2, \dots, b_{n-s}$; by definition, each blue edge together with certain red edges forms exactly one cycle, denoted respectively as $D_1, D_2, \dots, D_{n-s}$.

We first prove that there is no $D_i = D_{j_1} \star D_{j_2} \star \dots \star D_{j_k}$, because each blue edge belongs uniquely to one $D_i$, so as long as $i \neq j_1, j_2, \dots, j_k$, the above equation cannot hold.

Next, we prove that any other cycle can be obtained by applying the *-operation to some of the cycles among $D_1, D_2, \dots, D_{n-s}$. First, suppose there is a cycle containing $m$ blue edges (in order along the cycle, $b_{l_1}, \dots, b_{l_m}$). Let $D_i = E_i + F_i + b_i$, where $E_i, F_i$ are respectively the sets of edges from the two endpoints of $b_i$ to the root of the tree. Then
$$D_{l_1} * D_{l_2} * \dots * D_{l_m} = (E_{l_1} * F_{l_1} * b_{l_1}) * \dots * (E_{l_m} * F_{l_m} * b_{l_m}) = b_{l_1} * (E_{l_1} * F_{l_2}) * b_{l_2} * (E_{l_2} * F_{l_3}) * \dots * b_{l_m} * (E_{l_m} * F_{l_1})$$
where $E_{l_k} * F_{l_{k+1}}$ is the red segment between $b_{l_k}$ and $b_{l_{k+1}}$, hence $D_{l_1} * D_{l_2} * \dots * D_{l_m}$ is exactly the desired cycle.

Finally, we prove that $r = n - s$. In fact, if $r \ge n - s + 1$, then from these $r$ cycles we select $n - s + 1$ of them, denoted $G_1, G_2, \dots, G_{n-s+1}$. From the previous discussion we know that these cycles can all be obtained by applying the *-operation to some of the cycles among $D_1, D_2, \dots, D_{n-s}$. Therefore, if we replace addition with the *-operation, this can be expressed as:
$$X \begin{pmatrix} D_1 \\ D_2 \\ \vdots \\ D_{n-s} \end{pmatrix} = \begin{pmatrix} G_1 \\ G_2 \\ \vdots \\ G_{n-s+1} \end{pmatrix} \quad (1)$$
where $X$ is a $(n-s+1) \times (n-s)$ 0-1 matrix, and performing Gaussian elimination by adding and subtracting rows of $X$, there must be a row that becomes entirely 0 (i.e., row echelon form), meaning that one of the rows can be obtained by applying the *-operation to some of the other rows. That is, among $G_1, G_2, \dots, G_{n-s+1}$ there is also some element that can be obtained by applying the *-operation to the other elements among them, a contradiction. Therefore $r = n-s$, that is, $n = r+s$.