$ \Leftrightarrow 2^n|3999^n + 4001^n$ The first we has $ n$ is odd . Consider $ n\equiv 1(\mod 2)$ Then we has : $ 3999^n + 4001^n = 8000(\frac {3999^n + 4001^n}{8000})$ Lemma: $ \gcd(a + b,\frac {a^n + b^n}{a+b}) = \gcd(n,a + b)$ where $ \gcd(a,b) = 1$ So $ \gcd(8000,(\frac {3999^n + 4001^n}{8000}) = \gcd(800,n)$ So $ 2^n|8000$ so $ n\in = {1,3,5\}}$ Result There are 3 number in this sequence is interger.

Your solution looks correct, but can you please add a proof of the lemma in your post?

Yes it from Newton express. $ \gcd(a+b,\frac{a^n+b^n}{a+b})=\gcd(a+b,na^n)=\gcd(n,a+b)$

We can rephrase the problem as: Solve for $n$ such that $$v_2(3999^n+4001^n) \ge n.$$ Note that if $n$ is an odd integer, we have $$3999^n+4001^n=(3999+4001)*(3999^{n-1}4001+3999^{n-2}4001^2 \dots +4001^{n-1}).$$Clearly the latter part of the expansion is not divisible by $2$, because modulo $2$ on gives $1+1 \dots +1$ iterated $n$ times, which is odd. Now computing gives $v_2(3999^n+4001^n)=6v_2(n)=6$. This gives the solutions $n=1,3,5$. For evens, we can simply note that for even $n$, $2$ trivially divides $3999^n+4001^n$ but $4$ does not divide $3999^n+4001^n$, as modulo $4$ gives $(-1)^n+(1)^n \equiv 2 \pmod{4}$. Thus no even $n$ holds.