The squares of the one-digit positive integers $1,2,3,4,5,6,7,8,9$ are $1,4,9,16,25,36,49,64,81$, respectively.
Of these, the squares $1,25,36$ end with the digit of their square root.
In other words, $k=1,5,6$ are Anderson numbers.
Thus, $k=6$ is the only even one-digit Anderson number.
To find all even two-digit Anderson numbers, we note that any two-digit even Anderson number $k$ must have a units (ones) digit of 6. This is because the units digit of $k$ and the units digit of $k^{2}$ must match (by the definition of an Anderson number) and because the units digit of $k$ completely determines the units digit of $k^{2}$. (We can see this by doing "long multiplication".) So we need to look for two-digit Anderson numbers $k$ with digits $c 6$.
Another way of writing the number $c 6$ is $k=10 c+6$. (This form uses the place values associated with the digits.)
In this case, $k^{2}=(10 c+6)^{2}=(10 c+6)(10 c+6)=(10 c)^{2}+6(10 c)+10 c(6)+6^{2}=100 c^{2}+120 c+36$. Note that $k^{2}=100\left(c^{2}+c\right)+10(2 c+3)+6$ and so the units digit of $k^{2}$ is 6.
For $k$ to be an Anderson number, we need the tens digit of $k^{2}$ to be $c$, in which case the final two digits of $k^{2}$ will be $c 6$.
Thus, the tens digit of $k^{2}$ is equal to the units digit of $2 c+3$.
This means that $k=10 c+6$ is an Anderson number exactly when the units digit of $2 c+3$ is equal to the digit $c$.
When we check the nine possible values for $c$, we find that the only possibility is that $c=7$.
This means that $k=76$ is the only two-digit even Anderson number.
Note that $76^{2}=5776$, which ends with the digits 76.
Next, we look for three-digit even Anderson numbers $k$.
Using a similar argument to above, we see that $k$ must have digits $b 76$.
In other words, $k=100 b+76$ for some digit $b$.
In this case, $k^{2}=(100 b+76)^{2}=10000 b^{2}+15200 b+5776$.
We note that the tens and units digits of $k^{2}$ are 76, which means that, for $k$ to be an Anderson number, the hundreds digit of $k^{2}$ must be $b$.
Now $k^{2}=1000\left(10 b^{2}+15 b+5\right)+100(2 b+7)+76$.
Thus, $k$ is an Anderson number exactly when the units digit of $2 b+7$ is equal to the digit $b$.
Again, checking the nine possible values for $b$ shows us that $b=3$ is the only possibility.
This means that $k=376$ is the only three-digit even Anderson number.
Note that $376^{2}=141376$, which ends with the digits 376.
Since Anderson numbers are less than 10000, then we still need to look for four-digit even Anderson numbers.
Again, using a similar argument, we see that $k$ must have digits $a 376$.
In other words, $k=1000 a+376$ for some digit $a$.
In this case, $k^{2}=(1000 a+376)^{2}=1000000 a^{2}+752000 a+141376$.
We note that the hundreds, tens and units digits of $k^{2}$ are 376, which means that, for $k$ to be an Anderson number, the thousands digit of $k^{2}$ must be $a$.
Now $k^{2}=10000\left(100 a^{2}+75 a+14\right)+1000(2 a+1)+376$.
Thus, $k$ is an Anderson number exactly when the units digit of $2 a+1$ is equal to the digit $a$. Again, checking the nine possible values for $a$ shows us that $a=9$ is the only possibility.
This means that $k=9376$ is the only four-digit even Anderson number.
Note that $9376^{2}=87909376$, which ends with the digits 9376.
Thus, $S$, the sum of the even Anderson numbers, equals $6+76+376+9376=9834$.
The sum of the digits of $S$ is $9+8+3+4=24$.