Solution:
We show that infinitely many odd numbers in the sequence are divisible by $3$.

If $n = 14 \pmod{36}$, then $n = 36m + 14$ for some $m$. So there are $18m + 7$ odd numbers in the sum and $18m + 7$ even numbers. Hence the sum is odd.

There are $12m + 5$ numbers equal to $1 \pmod{3}$, $12m + 5$ equal to $-1 \pmod{3}$ and $12m + 4$ equal to $0 \pmod{3}$.

Any product of numbers equal to $1 \pmod{3}$ equals $1 \pmod{3}$, so if $k = 1 \pmod{3}$, then $k^{k} = 1 \pmod{3}$. Similarly, if $k = 0 \pmod{3}$, then $k^{k} = 0 \pmod{3}$. If $k = -1 \pmod{3}$ then $k^{k} = -1 \pmod{3}$ if $k$ is odd and $1$ mod if $k$ is even.

Of the $12m + 5$ numbers equal to $-1 \pmod{3}$, $6m + 3$ are even and $6m + 2$ are odd. Hence the sum $= (12m + 5) \cdot 1 + (6m + 3) \cdot 1 + (6m + 2) \cdot (-1) + (12m + 4) \cdot 0 = 12m + 6 \equiv 0 \pmod{3}$.

So the sum is divisible by $3$.