Andreas, Boyu, Callista, and Diane each randomly choose an integer from 1 to 9, inclusive. Each of their choices is independent of the other integers chosen and the same integer can be chosen by more than one person. The probability that the sum of their four integers is even is equal to $\frac{N}{6561}$ for some positive integer $N$. What is the sum of the squares of the digits of $N$ ?
Suppose that Andreas, Boyu, Callista, and Diane choose the numbers $a, b, c$, and $d$, respectively. There are 9 choices for each of $a, b, c$, and $d$, so the total number of quadruples $(a, b, c, d)$ of choices is $9^{4}=6561$.

Among the 9 choices, 5 are odd $(1,3,5,7,9)$ and 4 are even $(2,4,6,8)$.

If there are $N$ quadruples $(a, b, c, d)$ with $a+b+c+d$ even (that is, with the sum of their choices even), then the probability that the sum of their four integers is even is $\frac{N}{6561}$, which is in the desired form.

Therefore, we count the number of quadruples $(a, b, c, d)$ with $a+b+c+d$ even.

Among the four integer $a, b, c, d$, either $0,1,2,3$, or 4 of these integers are even, with the remaining integers odd.

If 0 of $a, b, c, d$ are even and 4 are odd, their sum is even.

If 1 of $a, b, c, d$ is even and 3 are odd, their sum is odd.

If 2 of $a, b, c, d$ are even and 2 are odd, their sum is even.

If 3 of $a, b, c, d$ are even and 1 is odd, their sum is odd.

If 4 of $a, b, c, d$ are even and 0 are odd, their sum is even.

Therefore, we need to count the number of quadruples $(a, b, c, d)$ with 0,2 or 4 even parts.

If 0 are even and 4 are odd, there are 5 choices for each of the parts, and so there are $5^{4}=625$ such quadruples.

If 4 are even and 0 are odd, there are 4 choices for each of the parts, and so there are $4^{4}=256$ such quadruples.

If 2 are even and 2 are odd, there are 4 choices for each of the even parts and 5 choices for each of the odd parts, and 6 pairs of locations for the even integers $(a b, a c, a d, b c, b d, c d)$ with the odd integers put in the remaining two locations after the locations of the even integers are chosen. Thus, there are $4^{2} \cdot 5^{2} \cdot 6=2400$ such quadruples.

In total, this means that there are $625+256+2400=3281$ quadruples and so $N=3281$.

The sum of the squares of the digits of $N$ is equal to $3^{2}+2^{2}+8^{2}+1^{2}=9+4+64+1=78$.

ANSWER: 78