Vietnamese MO, 2021

a) Notice that if $k$ is a positive integer such that $\gcd(k, n) = 1$ and $k < n$ then $n-k$ and $n$ are relatively prime. It follows that
$$\sum_{k \le n, \gcd(k,n)=1} k = \sum_{k \le n, \gcd(k,n)=1} (n-k)$$
Let $A = \{k \in \mathbb{N} \mid 1 \le k \le n, \gcd(k, n) = 1 = \{k_1, k_2, \dots, k_n\}\}$ then
$$\sum_{i=1}^{\varphi(n)} k_i = \frac{1}{2} \sum_{i=1}^{\varphi(n)} k_i + \frac{1}{2} \sum_{i=1}^{\varphi(n)} (n - k_i) = \frac{1}{2} \sum_{i=1}^{\varphi(n)} (k_i + n - k_i) = \frac{n\varphi(n)}{2}$$
So, we have
$$s(n) = \sum_{i=1}^{n} i - \sum_{i=1}^{\varphi(n)} k_i = \frac{n(n+1)}{2} - \frac{n\varphi(n)}{2} = \frac{n}{2}(n+1-\varphi(n)).$$

b) Suppose that there exists a positive integer $n$ such that $s(n) = s(n + 2021)$. Thus,
$$2s(n) = n(n + 1 - \varphi(n)) = (n + 2021)(n + 2022 - \varphi(n + 2021)) \quad (*)$$
or
$$2021(2n + 2022 - \varphi(n + 2021)) = n(\varphi(n + 2021) - \varphi(n)) \quad (1)$$
From (*), it follows that $n, n + 2021$ are divisors of $2s(n)$. Otherwise,
$$2s(n) = n(n + 1) - n\varphi(n) < n(n + 1) < n(n + 2021),$$
so $n, n + 2021$ are not coprime, which means $(n, 2021) \neq 1$. Let $d = \gcd(n, 2021) > 1$, so $\gcd(\frac{n}{d}, \frac{2021}{d}) = 1$. From (1), we get
$$\frac{2021}{d}(2n + 2022 - \varphi(n + 2021)) = \frac{n}{d}(\varphi(n + 2021) - \varphi(n)) \quad (2)$$
so there exists some positive integer $x$ such that
$$\varphi(n + 2021) - \varphi(n) = \frac{2021}{d} \cdot x \quad (3)$$
$$2n + 2022 - \varphi(n + 2021) = \frac{n}{d} \cdot x \quad (4)$$
Thus, $\varphi(n + 2021) - \varphi(n)$ is divisible by $2021/d$. Otherwise, one can check that $\varphi(n + 2021), \varphi(n)$ is divisible by $\varphi(d)$ and $\gcd(\varphi(d), 2021/d) = 1$ for all $d \in \{43, 47, 2021\}$, so
$$\varphi(n + 2021) - \varphi(n) : \frac{2021\varphi(d)}{d}$$
On the other hand, it follows from (3) and (4)
$$d < x = d \cdot \frac{2n + 2022 - \varphi(n)}{n + 2021} < 2d$$
It implies that, for all $d \in \{43, 47, 2021\}$
$$\frac{2021\varphi(d)}{d} < \frac{2021 \cdot d}{d} < \frac{2021}{d} \cdot x < 2 \cdot 2021 < \frac{3 \cdot 2021\varphi(d)}{d}$$
Thus, $\varphi(n + 2021) - \varphi(n) = 2 \cdot 2021\varphi(d)/d$ and $x = 2\varphi(d)$, so
$$\varphi(n + 2021) = \frac{2n(d - \varphi(d))}{d} + 2022 \quad (5)$$
$$\varphi(n) = \frac{2n(d - \varphi(d))}{d} + 2022 - \frac{2 \cdot 2021\varphi(d)}{d} \quad (6)$$
If $n$ has at most 10 distinct prime divisors then
$$\varphi(n) > n \prod_{i=2}^{11} \left(1 - \frac{1}{i}\right) = \frac{n}{11} > \frac{2n(d - \varphi(d))}{d}, d \in \{43, 47, 2021\},$$
which contradicts (6). We get $n$ has at least 11 distinct prime divisors, so $n > 12!$ and $\varphi(n)$ is divisible by $2^{10}$.
Similarly, if $n + 2021$ has at most 4 distinct prime divisors then
$$\varphi(n + 2021) > (n + 2021) \prod_{i=2}^{5} \left(1 - \frac{1}{i}\right) = \frac{n + 2021}{5}.$$
Otherwise, $n > 12!$ and
$$\frac{n + 2021}{5} > \frac{2n(d - \varphi(d))}{d} + 2022, d \in \{43, 47, 2021\},$$
which contradicts (5). We get $n + 2021$ has at least 5 distinct prime divisors, so $\varphi(n + 2021)$ is divisible by $2^4$.
On the other hand
$$v_2(\varphi(n + 2021) - \varphi(n)) = v_2\left(\frac{2 \cdot 2021\varphi(d)}{d}\right) \le 3, \forall d \in \{43, 47, 2021\}$$
which contradicts $\varphi(n + 2021):2^4$ and $\varphi(n):2^{11}$. Hence, there does not exist a positive integer $n$ such that $s(n) = s(n + 2021)$ □