20. Find the largest possible value of $n$ such that there exist $n$ consecutive positive integers whose sum is equal to 2010 .
20. Answer: 60
Let $a$ be a positive integer such that the sum of $n$ consecutive integers $a, a+1, \ldots$, $a+(n-1)$ is 2010 , that is,
$$
a+(a+1)+\cdots+(a+n-1)=2010 \text {. }
$$

This gives $\frac{n(2 a+n-1)}{2}=2010$, or
$$
n(2 a+n-1)=4020=2^{2} \times 3 \times 5 \times 67 .
$$

Since $n<2 a+n-1$, we have
$$
n<\sqrt{2^{2} \times 3 \times 5 \times 67}=2 \sqrt{1005}<2 \times 32=64 \text {. }
$$

Now (1) and (2) imply that $n \in\{1,2,3,4,5,6,10,12,15,20,30,60\}$. Since $n$ and $2 a+n-1$ have different parities, it follows that $n$ and $\frac{4020}{n}$ have different parities. Consequently, we have
$$
n \in\{1,3,4,5,12,15,20,60\} \text {. }
$$

If $n=60$, then $2 a+n-1=\frac{4020}{60}=67$, so $a=4$. Thus the largest possible value of $n$ is 60 .