已知等差数列\(\left\{ {{a_n}} \right\}\)的前\(n\)项和为\({S_n}\)，满足\({S_{19}} > 0\)，\({S_{20}} < 0\)，若数列\(\left\{ {{a_n}} \right\}\)满足\({a_m} \cdot {a_{m + 1}} < 0\)，则\(m=\)（       ）

等差数列\(\left\{ {{a_n}} \right\}\)的前n项和为\({S_n}\)，则\({S_{19}} = \frac{{{a_1} + {a_{19}}}}{2} \times 19 = 19{a_{10}} > 0\)，有\({a_{10}} > 0\)，

\({S_{20}} = \frac{{{a_1} + {a_{20}}}}{2} \times 20 = 10({a_{10}} + {a_{11}}) < 0\)，有\({a_{11}} < - {a_{10}} < 0\)，显然数列\(\left\{ {{a_n}} \right\}\)是递减的，且\({a_{10}} \cdot {a_{11}} < 0\)，

因\({a_m} \cdot {a_{m + 1}} < 0\)，所以\(m = 10\).

故选：B

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