1. Let $N=25^{190}-19^{190}-8^{190}+2^{190}.$ We need to check divisibility by $14$ and by $34$.

2. Check divisibility by $14$. Since $14=2\cdot 7$, it is enough to check modulo $7$ because $N$ is clearly even:

• $25^{190}$ is odd
• $19^{190}$ is odd
• $8^{190}$ is even
• $2^{190}$ is even So odd $-$ odd $-$ even $+$ even $=$ even.

Now reduce modulo $7$: $25\equiv 4 \pmod 7,\quad 19\equiv 5 \pmod 7,\quad 8\equiv 1 \pmod 7,\quad 2\equiv 2 \pmod 7.$ Thus $N\equiv 4^{190}-5^{190}-1^{190}+2^{190}\pmod 7.$ Now, $4^3=64\equiv 1\pmod 7 \Rightarrow 4^{190}=4^{3\cdot 63+1}\equiv 4,$ $5\equiv -2\pmod 7 \Rightarrow 5^{190}\equiv (-2)^{190}=2^{190},$ so those two terms cancel. Also, $2^3=8\equiv 1\pmod 7 \Rightarrow 2^{190}=2^{3\cdot 63+1}\equiv 2.$ Hence $N\equiv 4-2-1+2=3\pmod 7.$ So $N$ is not divisible by $7$, hence not divisible by $14$.

3. Check divisibility by $34$. Since $34=2\cdot 17$, again $N$ is even, so it remains to check modulo $17$.

Reduce modulo $17$: $25\equiv 8 \pmod{17},\quad 19\equiv 2 \pmod{17},\quad 8\equiv 8 \pmod{17},\quad 2\equiv 2 \pmod{17}.$ Therefore $N\equiv 8^{190}-2^{190}-8^{190}+2^{190}=0\pmod{17}.$ So $N$ is divisible by $17$. Since $N$ is also even, it follows that $N$ is divisible by $34$.

4. Conclusion:

• Not divisible by $14$
• Divisible by $34$

Therefore the correct option is $\boxed{\text{D: 34 but not by 14}}.$

5. Comparison with stored answer: Stored correct answer is D, which matches our result.