Four, (15 points) Can 2010 be written as the sum of squares of $k$ distinct prime numbers? If so, try to find the maximum value of $k$; if not, please briefly explain the reason.
If 2010 can be written as the sum of squares of $k$ prime numbers, taking the sum of the squares of the smallest 10 distinct primes, then
$$
\begin{array}{l}
4+9+25+49+121+169+ \\
289+361+529+841 \\
=2397>2010 .
\end{array}
$$

Therefore, $k \leqslant 9$.
It is easy to see that there is only one even prime number 2, and the rest are odd. The square of an odd number is congruent to 1 modulo 8.

Since 2010 is congruent to 2 modulo 8, but the sum of the squares of nine different primes is congruent to 1 or 4 modulo 8, and the sum of the squares of eight different primes is congruent to 0 or 3 modulo 8, hence $k \leqslant 7$.
When $k=7$, after trial calculation, we get
$$
2^{2}+3^{2}+7^{2}+11^{2}+13^{2}+17^{2}+37^{2}=2010 \text {. }
$$

In summary, 2010 can be written as the sum of squares of $k$ distinct prime numbers, and the maximum value of $k$ is 7.
In fact, it can also be proven that $k \neq 1,2, \cdots, 6$.
Therefore, 2010 can only be expressed as the sum of squares of seven distinct prime numbers.