Given that the sum of 3 distinct natural numbers is 55, and the sum of any two of these numbers is a perfect square, then these three natural numbers are
【Analysis and Solution】
Three natural numbers are all different, so the three sums of any two of them, which are perfect squares, are also different;
Adding the three perfect squares is equivalent to twice the sum of these three natural numbers;
Therefore, decompose $55 \times 2=110$ into the sum of three perfect squares;
The perfect squares not exceeding 110 are the following 11: $0^{2}=0, 1^{2}=1, 2^{2}=4, 3^{2}=9, 4^{2}=16, 5^{2}=25, 6^{2}=36, 7^{2}=49, 8^{2}=64, 9^{2}=81, 10^{2}=100$; $110 \div 3=36 \frac{2}{3}$, so there must be a perfect square greater than $36 \frac{2}{3}$;
Since the sum of the three different natural numbers is 55, these three perfect squares do not exceed 55; hence, one of the perfect squares must be $7^{2}=49$;
$110=49+61=49+36+25$, that is, these three perfect squares are $49, 36, 25$; $55-49=6, 55-36=19, 55-25=30$, that is, these three natural numbers are $6, 19, 30$.