Four different natural numbers, one of which is 1, the sum of any two is a multiple of 2, the sum of any three is a multiple of 3, and the sum of these four numbers is exactly a multiple of 4. The smallest sum of these 4 numbers is $\qquad$.

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Four different natural numbers, one of which is 1, the sum of any two is a multiple of 2, the sum of any three is a multiple of 3, and the sum of these four numbers is exactly a multiple of 4. The smallest sum of these 4 numbers is $\qquad$.
【Analysis and Solution】
Number Theory.
Let the 4 numbers be $1, a, b, c$;
Since the sum of $1+a$ is a multiple of 2; thus $a \div 2$ has a remainder of 1;
Similarly, $b \div 2$ has a remainder of 1, and $c \div 2$ has a remainder of 1;
Since $1+b+c$ is a multiple of 3, and $a+b+c$ is also a multiple of 3; thus 1 and $a$ are congruent modulo 3, i.e., $a \div 3$ has a remainder of 1; similarly, $b \div 3$ has a remainder of 1, and $c \div 3$ has a remainder of 1;

A number that $\div 2$ has a remainder of 1 and $\div 3$ has a remainder of 1; then this number $\div 6$ has a remainder of 1;
The smallest four natural numbers that satisfy $\div 6$ with a remainder of 1 are $1, 7, 13, 19$;
$1+7+13+19=40$, the sum of the four numbers is a multiple of 4;
Therefore, the smallest sum of these 4 numbers is 40.