7. (5 points) Given $m>1, m$ consecutive natural numbers sum up to 33, then the possible values of $m$ are $\qquad$
【Analysis】Since $m>1$, then the minimum value of $m$ is 2. We only need to find out the maximum value of $m$ to know the possible values of $m$; and when $m$ is the largest, it must start from 1, $1+2+3+4+5+6+7=2833$; so the maximum value of $m$ is 7. Therefore, all possible values of $m$ are the natural numbers from 2 to 7; then verify the values of $m$.

【Solution】Solution: $33 \div 2=16.5$, so $16+17=33$, thus $m=2$
$33 \div 3=11$ leads to $10+11+12=33$, thus $m=3$
$33 \div 4=8.25$ leads to $7+8+9+10=34$ so discard $m=4$
$33 \div 5=6.6$ leads to $5+6+7+8+9=35$ so discard $m=5$
$33 \div 6=5.5$ leads to $3+4+5+6+7+8=33$, thus $m=6$,
$33 \div 7=4.7,1+2+3+4+5+6+7+8=28<$ does not meet the requirement, discard $m=7$,
The sum of 8 consecutive natural numbers is definitely greater than 33, so $m$ cannot be a number greater than 8. Therefore, all possible values of $m$: $2, 3, 6$;

Hence the answer is: $2, 3, 6$.