We will say that two people are of the same type if they are both knights or both knaves, and of different type otherwise.
We note that, if person $A$ points to person $B$ and declares her a knight, then $A$ and $B$ are of the same type: both knights (if $A$ tells the truth) or both knaves (if $A$ lies). Conversely, if $A$ declares $B$ a knave, this means that $A$ is a knight and $B$ a knave, or $A$ a knave and $B$ a knight: $A$ and $B$ are of different type.

Let us number the people from $1$ to $n$ according to their order in the line, establishing that person $1$ is the one who is furthest ahead (and does not see anyone in front of her). We can now deduce from the declarations, for each person from the second to the $n$-th, whether she is or is not of the same type as person number $1$; in particular, the declaration of $2$ allows us to determine this for $2$, the declarations of $2$ and $3$ together determine it for $3$, and thus the declarations of the people from the permanent second to the $k$-th determine whether the latter is or is not of the same type as the first. The reason is the following: person $2$ necessarily points to the first, and is of her same type if she declares her a knight, of different type otherwise. Person $3$ points to $1$ or $2$: for both of them, thanks to the statement of $2$, we know whether their type is the same as $1$'s, and consequently we can deduce it for $3$. We proceed in the same way, in order, up to person $n$.

Suppose there are $m$ people of the type of $1$ (person number $1$ included) and $n-m$ people of the other type. We know that there must be more knaves than knights, so $m$ will be strictly greater than or strictly less than $n-m$.

In the first case we can deduce that $1$ and all the people of her type are knaves, the others knights; conversely in the second case.