Showing that minimal vertex deletion to a bipartite graph is NP-complete
Your problem is a special case of a broader class of problems named node-deletion problems:

J. M. Lewis and M. Yannakakis, "The node-deletion problem for hereditary properties is NP-complete"  

... This paper deals with the class of graph problems defined as follows:
for a fixed graph property $\Pi$, find the minimum number of nodes (or vertices) which must be deleted from a given graph $G$ so that the result satisfies $\Pi$. We call this the node-deletion problem for $\Pi$. Our results show that if $\Pi$ is a nontrivial property which is hereditary on induced subgraph, then the node-deletion problem for $\Pi$ is NP-hard. Furthermore, if we add the condition that testing for $\Pi$ can be performed in polynomial time, then our results imply that the node-deletion problem for $\Pi$ is NP-complete. ...

Your problem is the node deletion problem for bipartiteness, but (as noted by Pal), it is known today as the Odd cycle traversal (OCT) problem.

EDIT

For what regards a direct reduction, I thought of this one from 3SAT.

Given an instance of 3SAT with $n$ variables and $m$ clauses, build the following graph: add two nodes $x_i, \overline{x_i}$ for each variable and an edge between them. To simulate a truth assignment, add $n+1$ nodes for each variable $x_i$ and connect them both to $x_i$ and $\overline{x_i}$; in this way, in order to make a bipartite graph deleting at most $n$ nodes, at least one between $x_i$ and $\overline{x_i}$ must be deleted . Finally for each clause $C_j$ add 4 nodes and build an odd cycle that connects the variables in $C_j$.

The resulting graph $G$ can be made bipartite deleting at most $n$ nodes if and only if the original 3SAT formula is satisfiable.