Let's determine those four-digit numbers divisible by 28, which can be expressed as the sum of the squares of three consecutive even numbers.
Let the middle of three consecutive even numbers be denoted by $k$. Then the sum of the squares of these three even numbers is:

$$
(2 k-2)^{2}+(2 k)^{2}+(2 k+2)^{2}=12 k^{2}+8
$$

We want this number to be a four-digit number and divisible by 28.

a) Since $12 k^{2}+8$ is a four-digit number, its value must lie between 1000 and 9999 (inclusive):

$$
1000 \leqq 12 k^{2}+8 \leqq 9999
$$

from which

$$
10 \leqq k \leqq 28
$$

b) Since $12 k^{2}+8=4\left(3 k^{2}+2\right)$, the desired number is divisible by 28 if and only if $3 k^{2}+2$ is divisible by 7. Let's examine the remainders when divided by 7!

| $k$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | $9 \ldots$ |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| the remainder | 2 | 5 | 0 | 1 | 1 | 0 | 5 | 2 | 5 | $0 \ldots$ |

The remainders form a 7-periodic pattern, repeating. This follows from the fact that the difference between the $(i+7)$-th and the $i$-th number is divisible by 7:

$$
\left(3(i+7)^{2}+2\right)-\left(3 i^{2}+2\right)=7(6 i+21)
$$

Thus, $3 k^{2}+2$ is divisible by 7 if and only if $k$ gives a remainder of 2 or 5 when divided by 7.

Comparing the results from a) and b), we are looking for integers between 10 and 28 that give a remainder of 2 or 5 when divided by 7. These are:

$$
7+5=12, \quad 14+2=16, \quad 14+5=19, \quad 21+2=23, \quad 21+5=26
$$

Thus, $k$ can only be one of these values. The corresponding four-digit numbers 1736, 3080, 4340, 6356, 8120 indeed satisfy the desired condition.

$$
\begin{gathered}
1736=22^{2}+24^{2}+26^{2} ; \quad 3080=30^{2}+32^{2}+34^{2} \\
4340=36^{2}+38^{2}+40^{2} ; \quad 6356=44^{2}+46^{2}+48^{2} \\
8120=50^{2}+52^{2}+54^{2}
\end{gathered}
$$