The 61st question: Find the maximum value of the positive integer $n$, such that for any simple graph of order $n$ with vertices $v_{1}, v_{2}, \ldots, v_{n}$, there always exist $n$ subsets $A_{1}, A_{2}, \ldots, A_{n}$ of the set $\{1,2, \ldots, 2020\}$, satisfying: $A_{i} \cap A_{j} \neq \varnothing$ if and only if $v_{i}$ is adjacent to $v_{j}$.
Question 61:
Proof: (1) First, prove that $n \leq 89$.
Suppose $n \geq 90$, consider the complete bipartite graph $G(C, D)$, where $C=\{v_1, v_2, \ldots, v_{45}\}$, $D=\{u_1, u_2, \ldots, u_{n-45}\}$, and assume that the $n$ subsets $A_1, A_2, \ldots, A_{45}, B_1, B_2, \ldots, B_{n-45}$ satisfy the conditions.

For any $i \in \{1, 2, \ldots, n\}$, since $A_1$ intersects non-trivially with each $B_j (1 \leq j \leq n-45)$, and any two different $B_j$ have an empty intersection, we have $|A_i| \geq n-45$. Since $A_1, A_2, \ldots, A_{45}$ are pairwise disjoint, we have $|\cup_{i=1}^{45} A_i| = \sum_{i=1}^{45} |A_i| \geq 45(n-45) \geq 45^2 > 2020$, which leads to a contradiction.
(2) Next, prove that when $n \leq 89$, for any $n$-vertex simple graph, there exist subsets $A_1, A_2, \ldots, A_n$ that satisfy the conditions. Use mathematical induction to prove a more general conclusion: for any $n$-vertex simple graph, there always exist $n$ subsets $A_1, A_2, \ldots, A_n$ of $\{1, 2, \ldots, s_n\}$ that satisfy the conditions, where $s_n = \left[\frac{n^2}{4}\right]$ (when $n=1$, $A_1$ can only be the empty set).

When $n=1$, the condition is not contradictory, so the conclusion holds; when $n=2$, let $A_1 = \{1\}$, and determine $A_2$ to be $\{1\}$ or the empty set based on whether $v_1$ and $v_2$ are adjacent, so the conclusion still holds.
Assume the conclusion holds for $n=k$, and we will prove it for $n=k+2$.
If no two vertices are adjacent, we can take all $A_i$ to be the empty set. Below, assume there exist adjacent vertices, without loss of generality, let $v_1$ and $v_2$ be adjacent. By the induction hypothesis, for the induced subgraph formed by the other $k$ vertices $v_i (3 \leq i \leq k+2)$, there exist $k$ subsets $A_i' (3 \leq i \leq k+2)$ of $\{1, 2, \ldots, s_k\}$ that satisfy the corresponding conditions. Let $A_1' = A_2' = \emptyset$.
Call positive integers greater than $s_k$ "new elements".

Since $v_1$ and $v_2$ are adjacent, add the new element $s_k + 1$ to $A_1'$ and $A_2'$. For any $v_i (3 \leq i \leq k+2)$, if $v_i$ is not adjacent to both $v_1$ and $v_2$, no new elements are needed; if $v_i$ is adjacent to both $v_1$ and $v_2$, take the smallest unused new element and add it to $A_1'$, $A_2'$, and $A_i'$; if $v_i$ is adjacent to one of $v_1$ and $v_2$, say $v_1$, take the smallest unused new element and add it to $A_1'$ and $A_i'$ but not to $A_2'$. In any case, each $v_i (3 \leq i \leq k+2)$ uses at most one new element. In total, at most $k+1$ (including the new element used by $A_1'$ and $A_2'$) different new elements are used.

After the above series of operations of adding new elements, let $A_1', A_2', \ldots, A_{k+2}'$ become $A_1, A_2, \ldots, A_{k+2}$. Then for any $i, j (1 \leq i < j \leq k+2)$, $A_i \cap A_j \neq \emptyset$ if and only if $v_i$ and $v_j$ are adjacent. Since no more than $k+1$ new elements are used, the largest element does not exceed $s_k + 1 + k = s_{k+2}$, so the conclusion holds for $n = k+2$.

Therefore, the conclusion holds for all positive integers $n$. In particular, when $n \leq 89$, since $S_n \leq 44 \times 45 < 2020$, there exist subsets $A_1, A_2, \ldots, A_n$ that satisfy the conditions.
In summary, the maximum value of $n$ is 89.