A stochastic process $\{X_i\}_{i=0}^\infty $ is $n$-Markov if
$$P(X_{t+n}|X_{t+n-1}, X_{t+n-2}, \cdots , X_{t}) = P(X_{t+n}|X_{t+n-1})$$
for any $t \ge 0$.
We would prove that
An $n$-Markov stochastic process must be $m$-Markov while is not necessarily $l$-Markov where $l > n > m$
N+1 to N
First, we prove an (n+1)-Markov stochastic process must be n-Markov.
Proof: Suppose $\{X_i\}_{i=0}^\infty$ is an $(n+1)$-Markov stochastic process. We have
$$P(X_{t+n}|X_{t+n-1}, X_{t+n-2}, \cdots, X_t) = P(X_{t+n} | X_{t+n-1})$$ for any $t \ge 0$, deriving
$$H(X_{t+n}|X_{t+n-1}, X_{t+n-2}, \cdots, X_t) = H(X_{t+n} | X_{t+n-1}).$$
Note that
$$\begin{align*} & H(X_{t+n}|X_{t+n-1}, X_{t+n-2}, \cdots, X_{t+1}, X_t) \\ \le & H(X_{t+n}|X_{t+n-1}, X_{t+n-2}, \cdots, X_{t+1}) \\ \le & \cdots \\ \le & H(X_{t+n} | X_{t+n-1}). \end{align*} $$
Therefore, these expressions have the same value which means
$$ \begin{align*} & I(X_{t+n};X{t}|X_{t+n-1}, X_{t+n-2}, \cdots, X_{t+1}) \\ = & H(X_{t+n}|X_{t+n-1}, X_{t+n-2}, \cdots, X_{t+1}) - H(X_{t+n}|X_{t+n-1}, X_{t+n-2}, \cdots, X_{t+1}, X_t) \\ = & 0. \end{align*} $$
Thus, $X_{t+n}$ and $X_t$ are independent given $X_{t+n-1}, \cdots , X_{t+1}$. Using
$$ \begin{align*} & P(X_{t+n}|X_{t+n-1}, X_{t+n-2}, \cdots, X_{t+1}, X_{t}) \\ =& \frac{P(X_{t+n}, X_t |X_{t+n-1}, X_{t+n-2}, \cdots, X_{t+1})}{P(X_t |X_{t+n-1}, X_{t+n-2}, \cdots, X_{t+1})} \\ =& P(X_{t+n}|X_{t+n-1}, X_{t+n-2}, \cdots, X_{t+1}), \end{align*} $$
iteratively, we could easily gain the conclusion.
Q.E.D.
N to N+1
Then we prove that there exists an n-Markov process that is not (n+1)-Markov.
The question is actually asking whether it is possible that knowing n preceding states gains no more information than knowing the exact preciding state, but knowing $(n+1)$ preciding states gains more information.
Proof: Consider a stationary stachastic process $\{X_i\}_{i=0}^\infty$ in which $X_i$ is i.i.d. Bernoulli trial with $p = \frac12$ for $i=0,1,\cdots,n$ and
$$X_{t+n+1} = (X_{t+n} + \cdots + X_t)\, \text{mod}\, 2 $$
for $t \ge 0 $.
$P(X_{t+n+1}|X_{t+n}, X_{t+n-1}, \cdots, X_{t+1}, X_{t})$ is determined by its $n+1$ preceding random variables. On the other hand, $P(X_{t+n+1}|X_{t+n}, X_{t+n-1}, \cdots, X_{t+1}) = 1/2$.
Q.E.D.