4. Natural numbers $a$, $b$, and $c$, greater than 2022, are such that $a+b$ is divisible by $c-2022$, $a+c$ is divisible by $b-2022$, and $b+c$ is divisible by $a-2022$. What is the greatest value that the number $a+b+c$ can take? (S. Berlov)
Answer: 2022.85. Solution. Lemma. For any natural numbers $d_{1}, d_{2}, d_{3}$, if $1 / d_{1}+1 / d_{2}+1 / d_{3}<2$, then $d$ does not exceed $1 / 3+1 / 3+1 / 4=11 / 12$. If $d_{1}=2$ and $d_{2}>3$, then $d$ does not exceed $1 / 2+1 / 4+1 / 5=19 / 20$. Finally, if $d_{1}=2$ and $d_{2}=3$, then $d$ does not exceed $1 / 2+1 / 3+1 / 7=41 / 42$.

Consider the number $N=a+b+c-2022$. It is divisible by $a-2022, b-2022$, and $c-2022$. Let $d_{1}=N /(a-2022), \quad d_{2}=N /(b-2022), \quad d_{3}=N /(c-2022)$. Then $N / d_{1}+N / d_{2}+N / d_{3}=N-4044$, from which $1 / d_{1}+1 / d_{2}+1 / d_{3}=1-4044 / N$. By the lemma, $4044 / N \geq 1 / 42$, i.e., $N \leq 2022 \cdot 84$, from which $a+b+c=N+2022 \leq 2022 \cdot 85$. Examples are obtained if we take $N=2022 \cdot 84: a-2022=N / 2$ $\Rightarrow a=2022 \cdot 43 ; b-2022=N / 3 \Rightarrow b=2022 \cdot 29 ; c-2022=N / 7 \Rightarrow c=2022 \cdot 13$.