Let $b_{n}=\lfloor n(\sqrt{2}-1)\rfloor$ and $c_{n}=\lfloor n(\sqrt{3}-1)\rfloor$. Clearly, $b_{n}+c_{n}=a_{n}-2n$ has the same parity as $a_{n}$ for all $n$, and also $b_{n}\le b_{n+1}\le b_{n}+1$ and likewise for $c_{n}$. It is easy to see (using the standard argument that multiples of an irrational have dense fractional parts) that there are infinitely many positive integers $k$ such that $\{ k(\sqrt{2}-1)\}<.1$ and $\{ k(\sqrt{3}-1)\}<.1$, where braces denote fractional part.

For any such integer $k$, note that $b_{k+2}=b_{k}$ and $c_{k+2}=c_{k}+1$, so $b_{k+2}+c_{k+2}=b_{k}+c_{k}+1$. Then $a_{k+2}$ and $a_{k}$ have different parity. Since there are arbitrarily large $k$, $a_{n}$ cannot be of constant parity after a certain point. Therefore there must be infinitely many evens and infinitely many odds among $a_{n}$.

Solution: I will use two well-known facts: 1st: \[ \sqrt {3} + \sqrt {2} \] is irrational number 2nd \[ \sqrt {3} - \sqrt {2} \] is irrational number,as well. Now let's return to the initial problem. Let's consider $ \sqrt {2}$ and $ \sqrt {3}$ in binary representation, $ \sqrt {2} = 1,b_{1}b_2b_3\dots$ $ \sqrt {3} = 1,c_{1}c_2c_3\dots$ Notice that if for some $ i$ we have that $ b_i + c_i = 1$,then $ [2^{i}\sqrt {2}] + [2^{i}\sqrt {3}]$ is obviously odd. And if $ b_i + c_i$ is even then $ [2^{i}\sqrt {2}] + [2^{i}\sqrt {3}]$ is even as well. But it is easy to show that there are infinitely many such numbers $ i$ of both type,otherwise we would obtain a contradiction to either fact no.1 or no.2 $ \blacksquare$.

It is equivalent to prove the same about $c_{n}=\lfloor n\sqrt 3 \rfloor-\lfloor n\sqrt 2 \rfloor$. Assume that the sequence $C=c_1,c_2,...$ contains only finite number of odd or even numbers. Since each two successive terms differ by at most 1, the sequence must become constant after some point. This is of course impossible, since the sequence is clearly unbounded.