The topology here treated is $\mathcal{T} = \{U \subset \mathbb{N} : n \in U, n \text{ odd } \Rightarrow n+1 \in U\}$. The aim is to guess if this space is locally connected or locally path connected (or both).
In relation with de local path connectedness, I think that the space is not locally path connected. I thought I could take an odd $n \in \mathbb{N}$ and say that every neighbourhood has to contain the basic open set $\{n, n+1\}$. So, if I proof that this set is not path connected, then I could afirm that none of the neighbourhoods are path connected. To proof this, I thought about trying to create a path begining on the $n+1$ and reaching to the conclusion that the path has to be constant.
On the other hand, I think that the space is locally connected. To proof this I though about taking the neighbourhood basis: $$B_n= \{n\}_{n=2k} \cup \{n, n+1\}_{n=2k+1}.$$ And proving that every set in that basis is connected.
The crucial observation is that if $X$ has the following property: for any $x\in X$ there is the smallest open neighbourhood $U_x$ of $x$ then $X$ is locally (path) connected if and only if each $U_x$ is (path) connected. And in fact most local properties translate to properties of each $U_x$.
And our space has this property:
If $n$ is even, then $U_n=\{n\}$ is obviously the smallest open neighbourhood and it is path connected.
If $n$ is odd, then $U_n=\{n,n+1\}$ is the smallest open neighbourhood of $n$. The (relative) topology on $U_n$ is given by $\tau=\big\{\emptyset, \{n+1\}, \{n,n+1\}\big\}$. It is the Sierpiński space. So is that space path connected? Since it has only two points then it is enough to show that we can join them via a path. Indeed, consider
$$f:I\to U_n$$ $$f(t)=\begin{cases} n &\text{if }t=0 \\ n+1 &\text{otherwise} \end{cases}$$
Can you verify that $f$ is a continous path from $n$ to $n+1$?
Therefore $\mathbb{N}$ with your custom topology is locally path connected.