1. If the sum of $k$ consecutive positive integers is 2010, then the maximum value of $k$ is
Ni. 1.60 .
Let $2010=(n+1)+(n+2)+\cdots+(n+k)$.
Then $k(2 n+k+1)=4020$.
Notice that $k<2 n+k+1$, and
$$
4020=2^{2} \times 3 \times 5 \times 67 \text {, }
$$

To maximize the value of $k$, 4020 should be expressed as the product of the closest pair of factors, which is $4020=60 \times 67$.
Thus, $k_{\max }=60$.