在数列 \(\left\{ {{a_n}} \right\}\)中,\({a_1} = \frac{1}{2}\),\({a_{n + 1}} = 1 - \frac{1}{{{a_n}}}\),则 \({a_2} = 1 - \frac{1}{{{a_1}}} = - 1\),\({a_3} = 1 - \frac{1}{{{a_2}}} = 2\),\({a_4} = 1 - \frac{1}{{{a_3}}} = \frac{1}{2}\),以此类推可知,对任意的 \(n \in {N^ * }\),\({a_{n + 3}} = {a_n}\),即数列 \(\left\{ {{a_n}} \right\}\) 是以\(3\)为周期的周期数列,
\(\because 2021 = 3 \times 673 + 2\), 因此,\({S_{2021}} = 673{S_3} + {a_1} + {a_2} = 674{S_3} - {a_3} = 674 \times \left( {\frac{1}{2} - 1 + 2} \right) - 2 = 1009\).
故选:B.