Question 225, Find the smallest positive integer $n(n \geq 4)$, such that from any $n$ different integers, one can always select four distinct numbers $a, b, c, d$, satisfying $20 \mid (a+b-c-d)$.
Question 225, Solution: First, consider the different residue classes modulo 20. Let $\mathrm{M}$ be a set of $\mathrm{k}$ elements that are mutually incongruent modulo 20. Define the set $A=\left\{a_{i}+a_{j} \mid a_{i}, a_{j} \in M, 1 \leq i < j \leq k\right\}$. Since there are more than 20 elements in $A$, by the pigeonhole principle, there must be two numbers $a+b$ and $c+d$ in $A$ that belong to the same residue class modulo 20, i.e., $20 \mid a+b-c-d$.

In general, consider a set $\mathrm{T}$ consisting of 9 different elements. If the elements of $\mathrm{T}$ belong to at least 7 different residue classes modulo 20, then according to the previous conclusion, there must exist four distinct elements $a, b, c, d$ in $\mathrm{T}$ such that $20 \mid a+b-c-d$. If the elements of $\mathrm{T}$ belong to at most 6 different residue classes modulo 20, then there must be 4 numbers that belong to the same residue class modulo 20, or two pairs of numbers that belong to the same residue class modulo 20. For both cases, we can select four distinct numbers $a, b, c, d$ such that $20 \mid (a+b-c-d)$.

On the other hand, for a set of 8 elements, we can take the set $\{0, 20, 40, 1, 2, 4, 7, 12\}$. The residues of these 8 elements modulo 20 are $0, 0, 0, 1, 2, 4, 7, 12$. These residues satisfy: each number is greater than the sum of any two smaller numbers, and the sum of any two numbers is less than 20, so there does not exist $a+b \equiv c+d (\bmod 20)$. This means that for any four distinct numbers $a, b, c, d$, we have $20 \nmid (a+b-c-d)$.
In summary, the smallest positive integer $\mathrm{n}$ that satisfies the condition is 9.