This argument is due to Burkhard Wilking. The other answers, answer to the OP question very well but I want to give a sharp example. I.e.
$$\sec>0 \implies\!\!\!\!\!\!\!\!\!\!\not\quad\ \ \mathcal{R}\geq0$$.
Any positive sectional curvature metric on $\mathsf{SU(3)}/\mathsf{T^2}$ has some negative curvature operator. To see this. note that it is an irreducible simply connected Riemannian manifold of positive sectional curvature and it is not diffeomorphic to any symmetric space; it follows that it should have some negative curvature operator by classification of closed manifolds of nonnegative curvature operator. See for instance Page 270, theorem 7.34 of
Chow, Bennett; Lu, Peng; Ni, Lei, Hamilton’s Ricci flow, Graduate Studies in Mathematics 77. Providence, RI: American Mathematical Society (AMS) (ISBN 0-8218-4231-5/hbk). xxxvi, 608 p. (2006). ZBL1118.53001.