If $G$ is an infinite graph where $v_{i,j}$ is joined with $v_{k,i+j}$ for all $k,i,j$ then is the chromatic number of G always infinite?

For every $n \in \mathbb{N}$, $G$ contains a copy of the $n$th shift graph, $G_n$. These are a classic construction of (finite) triangle-free graphs with arbitrarily large chromatic number. The vertices of $G_n$ are all intervals $[i,j]$ with $1 \leq i < j \leq n$. Two intervals $[i,j]$ and $[k,\ell]$ are adjacent if $j=k$ or $\ell=i$. It is well-known that $G_n$ has chromatic number $\lceil \log_2 n \rceil$.

Now, let $f:V(G_n) \to V(G)$ be defined by $f([i,j])=v_{j-i,i}$. Clearly, $f$ is an injective homomorphism, and so $G_n$ is a subgraph of $G$ for all $n$. Thus, $G$ has infinite chromatic number. This answers a question of Fedor Petrov asked in a comment to Ilya Bogdanov's answer. Actually, the proof that the shift graphs have unbounded chromatic number is the same argument given in Ilya's answer.

The chromatic number is indeed infinite.

Assume that there is a proper coloring in finitely many colors. Denote by $S_i$ the set of colors of the vertices having the form $v_{k,i}$ (for some $k$). There are two equal sets, say $S_i$ and $S_j$ with $i<j$. Then the color of $v_{j-i,i}$ lies in that set, and therefore is also a color of some vertex $v_{k,j}$; but those vertices are adjacent.