$3|2010$, but if $p_i\not =3, p_i^2=1\mod 3$. Therefore number of primes, $p_i$, suth that $p_i\not =3$ is $3m$. $2010=2\mod 8$. Therefore if $p_1\not =2$, then number of primes are 2 mod 8. Because $3|2010$ we get $k>2$. Minimal value for $k$ is $7$. In these case $p_1=2,p_2=3$. If $p_3>5$ then $5|\sum_ip_i^2$, only if $p_i^2=-1$ for 6 primes or $p_i\in \{7,13,17,23,37,43\}$. If $k>7$, then $k=10$ or $k=15$. In last case $p_1=2,p_2=5$ and $p_{15}\ge 47$ - contradition. when $k=10$ we get $p_1=3.$ Therefore only $k=6$ or $k=10$. But I did'nt chek.

Of course we've got $ 24|p^{2}-1$ for $ p>3 $. From that we have $ p_1=2 \ \ p_2=3 $ and only one possible value for k is 7. And then we should show example... -_- If I remember it should be $ 2^2+3^2+5^2+11^2+19^2+23^2+31^2=2010 $ Awful problem, I wonder how it could be on Baltic Way...

only solution k=7 example $2^2+3^2+5^2+7^2+11^2+29^2+31^2$ it's easy to prove that others cannot not hold using modulo 8 and a little inequality

actually,modulo 8 k=2 or k>10 if k=2,since 3||2010 it cannot be represented as sum of two squares so k>10 but 3^2+5^2+...+31^2=3354>2010 so p1=2 so $\sum p_i=2006$ modulo 8 then k=7 or k>15(contradiction).

Let wlog $p_1<p_2<\cdots <p_k$. Then $p_3^2\equiv p_4^2\equiv\cdots\equiv p_k^2\equiv 1\pmod{24}$, so $p_1^2+p_2^2\equiv 20-k\pmod{24}$. $k\le 9$, because if $k\ge 10$, then $\sum_{i=1}^k p_i^2=2010\ge 2^2+3^2+5^2+7^2+11^2+13^2+17^2+19^2+23^2+29^2=2397$, contradiction. If $p_1=2, p_2=3$, then $k\equiv 7\pmod{24}$, so $k=7$. Probably brute force is required here to find an example. $(p_1,p_2,p_3,p_4,p_5,p_6,p_7)=(2,3,5,7,11,29,31), (2,3,5,11,19,23,31),(2,3,7,11,13,17,37),(2,3,7,13,17,23,31)$ are all the solutions, but finding one is enough for this problem. If $p_1=2, p_2\neq 3$, then $k\equiv 15\pmod{24}$, no solutions. If $p_1=3$, then $k\equiv 10\pmod{24}$, no solutions. If $p_1>3$, then $k\equiv 18\pmod{24}$, no solutions.

Here we go: Naively, we can have $k \le 14$ at-max we can have $43^2=1849$ in the sum (which is 14th prime's square). If $p_i$ are all odd: Then take $\pmod 8$ (note that $p_i^2 \equiv 1 \pmod 8$). Thus: $k \equiv 2 \pmod 8$. Either $k=2$ or $k=10$. Case-1: $k=2$: Taking $\pmod 3$ gives us contradiction. Case-2: $k=10$ Notice that, $p_1^2 + p_2^2 + \cdots + p_10^2 \ge 2^2+3^2+5^2+7^2+11^2+13^2+17^2+19^2+23^2+29^2=2397$. Thus, we cant have $10$ primes (with a nice bound of $k \le 9$). If one of $p_i$ is $p_1=2$: Taking $\pmod 8$ again: we get $k \equiv 7 \pmod 8$. Thus: $k=7$. Taking $\pmod 3$ shows that one of the primes must be $p_2=3$. From here on-wards, we take leap of faith and hunt it down: Take one of the primes to be $p_3=5$. Taking $\pmod 5$ shows $3$ of the remaining primes should be $p^2 \equiv 1 \pmod 5$ and the remaining one with $p^2 \equiv -1 \pmod 5$. We choose $p_4 = 7$ to be the prime with $p^2 = 49 \equiv -1 \pmod 5$. After a bit more of hunting: $p_5=11, p_6=25, p_{7}=31$ works and we are done.