$1$.
$x+(x+1)+(x+2)+...+(x+30)=1240$
$⇔ 31x + (1+2+...+30) = 1240$
$⇔ 31x + \dfrac{(30+1).[(30-1):1+1]}{2} = 1240$
$⇔ 31x + 465 = 1240$
$⇔ 31x = 775$
$⇔ x = 25$
Vậy $x=25$
$2$.
$1+2+3+....+x = 210$
$⇔ \dfrac{(x+1).[(x-1):1+1]}{2} = 210$
$⇔ (x+1).x = 420$
$⇒$ \(\left[ \begin{array}{l}x=20\\x=-21(KTM)\end{array} \right.\)
Vậy $x=20$