We know $2023=7\times 289$

So ${\left(2021\right)}^{2022}+{\left(2022\right)}^{2021}$

$={\left(2023-2\right)}^{2022}+{\left(2023-1\right)}^{2021}$

$=7m_1+2^{2022}+7m_2+{\left(-1\right)}^{2021}$

$=7\left(m_1+m_2\right)+8^{674}-1$

$=7\left(m_1+m_2\right)+(7+1{)}^{674}-1$

$=7\left(m_1+m_2\right)+7m_3+1-1$

$=7\left(m_1+m_2+m_3\right)$

This is a multiple of $7$

Hence, remainder is $0$