$1+2+...+n=820$
$⇔\frac{(n+1)n}{2}=820$
$⇔n^2+n=1640$
$⇔n^2+n-1640=0$
$⇔n^2+n-1640=0$
$⇔(n^2-40n)+(41n-1640)=0$
$⇔(n-40)(n+41)=0$
$⇔\left[ \begin{array}{l}x-40=0\\x+41=0\end{array} \right.⇔\left[ \begin{array}{l}x=40(tm)\\x=-41 (loại)\end{array} \right.$
Vậy $n=40$.