Can we prove a stronger claim? Here
it is asked whether for every $k>1$, there is a prime $p$ and $w_1,w_2,\cdots ,w_k>1$, such that $p$ divides $w_1^2+w_2^2+\cdots +w_k^2$, but none of the summands. 

Can we prove there is a solution for every $k$, if the $w$s must be pairwise distinct ? The prime $p$ need not be the same for different $k$s.

 A: By the argument described here; we can prove a more stronger problem;
i.e. we can let $k \geq 3$ and $p\geq 7$ both be arbitrary. 

A: For every $k>1$, there is a prime $p$ and $w_1,w_2,\cdots ,w_k>1$, such that $p$ divides $w_1^2+w_2^2+\cdots +w_k^2$, but none of the summands.
Consider you have an ordered $k$-tuple $(w_1,w_2,\cdots ,w_k)$ satisfiying A; then for every $1 \leq i \leq k$ and every $k \in \mathbb{N}$ ; let 
$$w_i':=w_i+kp \ \ \text{  and for every  } \ \ j\neq i \ \ \ w_j':=w_j;$$ 
then one can checks that the ordered $k$-tuple $(w_1',w_2',\cdots ,w_k')$ will satisfies A too.

Doing this procedure consecutively we will get an ordered $k$-tuple with distinct elements.
A: Yes. Let $p$ be any prime dividing $k$ (it exists since $k > 1$). Then set $w_1 = p + 1, \: w_2 = 2p + 1, \: \dots, \: w_k = kp + 1$ (in general, $w_i = ip + 1$).
Now, clearly $p \nmid w_i$ and the $w_i$'s are pairwise distinct. Moreover, $$w_1^2 + w_2^2 + \cdots + w_k^2 \equiv \underset{k \: \text{times}}{\underbrace{1 + \cdots + 1}} = k \equiv 0 \pmod{p}$$ Thus $p \mid w_1^2 + \cdots + w_k^2$.
A: $$
\text{One more solution . . .}
\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;
$$
\begin{align*}
\text{Claim:}\;&\text{The prime $p=17$ works for all integers $k > 1$.}\\[8pt]
\text{Case$\;(1)\text{:}\;$}&\text{$k$ is even.}\\[8pt]
&\text{Choose integers}\;x_1,...,x_k,\;\text{all distinct, such that}\\[8pt]
&\qquad{\small\bullet}\;\;{\small{\frac{k}{2}}}\;\text{of them are}\;\equiv 4\;(\text{mod}\;17)\\[4pt]
&\qquad{\small\bullet}\;\;{\small{\frac{k}{2}}}\;\text{of them are}\;\equiv 1\;(\text{mod}\;17)
\\[8pt]
&\text{Then}\;17\;\text{doesn't divide any of $x_1,...,x_k$}\\[4pt]
&\text{but}\;\;\sum_{i=0}^k x_i^2 \equiv 
\left({\small{\frac{k}{2}}}\right)4^2 
+
\left({\small{\frac{k}{2}}}\right)1^2 
\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv\;
\left({\small{\frac{k}{2}}}\right)(4^2+1^2)\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv\;
\left({\small{\frac{k}{2}}}\right)(0)\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv\;0\;(\text{mod}\;17)\\[4pt]
&\text{which proves the claim for case $(1)$.}\\[8pt]
\text{Case$\;(2)\text{:}\;$}&\text{$k$ is odd, $k \ge 3$.}\\[8pt]
&\text{Choose integers}\;x_1,...,x_k,\;\text{all distinct, such that}\\[8pt]
&\qquad{\small\bullet}\;\;\text{one of them is}\;\equiv 6\;(\text{mod}\;17)\\[4pt]
&\qquad{\small\bullet}\;\;{\small{\frac{k+1}{2}}}\;\text{of them are}\;\equiv 4\;(\text{mod}\;17)\\[4pt]
&\qquad{\small\bullet}\;\;{\small{\frac{k-3}{2}}}\;\text{of them are}\;\equiv 1\;(\text{mod}\;17)
\\[8pt]
&\text{Then}\;17\;\text{doesn't divide any of $x_1,...,x_k$}\\[4pt]
&\text{but}\;\;\sum_{i=0}^k x_i^2 \equiv 
6^2 + \left({\small{\frac{k+1}{2}}}\right)4^2
+
\left({\small{\frac{k-3}{2}}}\right)1^2
\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv 
2 + \left({\small{\frac{k+1}{2}}}\right)16
+
\left({\small{\frac{k-3}{2}}}\right)1
\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv 
2 + \left({\small{\frac{k+1}{2}}}\right)16
+
\left({\small{\frac{k-3}{2}}}\right)18
\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv 
2 + (8k+8) + (9k-27)
\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv 
17k-17\;(\text{mod}\;17)\\[4pt]
&\qquad\qquad\;\;\;\;\equiv\;0\;(\text{mod}\;17)\\[4pt]
&\text{which proves the claim for case $(2)$.}\\[8pt]
&\text{Thus, the proof of the claim is complete.}\\[8pt]
\end{align*}
