We want the following property holds with a high probability:
$\forall i, 0\leq i \leq n: PRF(k_1, i)\neq PRF(k_2,i)$
You don't specify the sizes of the key space or codomain of the PRF. The latter in particular is relevant; let's call $t$ (for "tag") the size of the PRF's codomain. Modeling the PRF family as random functions, for any specific choice of $i$, the probability that the two functions collide is:
$$Prob\big[PRF(k_1, i) = PRF(k_2,i)\big] = {1\over{t}}$$
Negating this proposition, we get:
$$Prob\big[PRF(k_1, i) ≠ PRF(k_2,i)\big] = {1 - {1\over{t}}}$$
Now the problem is your universal quantifier: you're asking that there be no collision for any choice of $i \in [0, n]$. This means that we just need to find one colliding choice of $i$ to falsify it. So you're asking for the probability of this conjunction of independent events:
$$PRF(k_1,i_0) ≠ PRF(k_2,i_0) \land \dots \land PRF(k_1,i_n) ≠ PRF(k_2,i_n)$$
...which is the product of the probabilities of the conjuncts. Therefore:
$$Prob\big[\forall i, 0\leq i \leq n: PRF(k_1, i)\neq PRF(k_2,i)\big] = \left({1 - {1\over{t}}}\right)^{n+1}$$
This probability obviously falls exponentially as $n$ gets bigger. But you want a "pseudorandom" function that "maximizes" that probability. You didn't specify precisely what you mean by maximize, but I'd suggest that a function for which the probability doesn't go down exponentially with $n$ may not be pseudorandom. A function like what you're looking for does not behave like a random function, at least not for large values of $n$.
For small values of $n$, on the other hand, you may well find the probability high enough to your liking that a pseudorandom function is OK.