Problem 850

Lemma: In any $m$ integers, there must be some whose sum is a multiple of $\mathbf{m}$.
Proof: Let $\mathrm{m}$ integers be $\mathrm{x}_{1}, \mathrm{x}_{2}, \cdots, \mathrm{x}_{\mathrm{m}}, \mathrm{S}_{\mathrm{i}}=\mathrm{x}_{1}+\mathrm{x}_{2}+\cdots+\mathrm{x}_{\mathrm{i}}(\mathrm{i}=1,2, \ldots, \mathrm{m})$. If $\mathrm{S}_{1}, \mathrm{~S}_{2}, \cdots, \mathrm{S}_{\mathrm{m}}$ have distinct remainders when divided by $\mathrm{m}$, then there must be $\mathrm{m} \mid \mathrm{S}_{\mathrm{i}}$; otherwise, there must be $1 \leqslant \mathrm{i}<\mathrm{j} \leqslant \mathrm{m}$ such that $\mathrm{S}_{\mathrm{i}}=\mathrm{S}_{\mathrm{j}}$ (modm), then $\mathrm{m} \mid \mathrm{S}_{\mathrm{j}}-\mathrm{S}_{\mathrm{i}}=\mathrm{x}_{\mathrm{i}+1}+\cdots+\mathrm{x}_{\mathrm{j}}$, proving the lemma. Returning to the original problem.
$2020=2 \times 1010$, the sum of 2019 ones and 1 zero is $2019,2020 \nmid 2020-1$, so there does not exist an even number of 1s in 2019 ones and 1 zero whose sum is a multiple of 2020, hence $\mathrm{k} \geqslant 2021$.

Taking any 2021 integers, let there be $\mathrm{t}$ odd numbers $a_{1}, a_{2}, \ldots, a_{\mathrm{t}}$, and $\mathrm{s}$ even numbers $b_{1}, b_{2}, \ldots, b_{\mathrm{s}}$ where $\mathrm{t}+\mathrm{s}=2021$. If $\mathrm{t}$ is odd, then $\mathrm{s}$ is even. Let $x_{1}=\frac{a_{1}+a_{2}}{2}, x_{2}=\frac{a_{3}+a_{4}}{2}, \cdots, x_{\frac{t-1}{2}}=\frac{a_{t-2}+a_{t-1}}{2}$, $x_{\frac{t+1}{2}}=\frac{b_{1}+b_{2}}{2}, x_{\frac{t+1+1}{}+1}=\frac{b_{3}+b_{4}}{2}, \cdots, x_{1010}=\frac{b_{s-1}+b_{s}}{2}$. By the lemma, $x_{1}, x_{2}, \ldots, x_{1010}$ must have some whose sum is a multiple of 1010, i.e., $a_{1}, a_{2}, \ldots, a_{1}, b_{1}, b_{2}, \ldots, b_{\mathrm{s}}$ have an even number whose sum is a multiple of 2020. Therefore, $\mathrm{k} \leqslant 2021, \mathrm{k}=2021$. When $\mathrm{t}$ is even and $\mathrm{s}$ is odd, similarly $\mathrm{k}=2021$. In summary, $\mathrm{k}=2021$.