函数 \(f(x) = \frac{1}{{x + 1}}\),当 \(x > 0\) 时,\(f(x) + f(\frac{1}{x}) = \frac{1}{{1 + x}} + \frac{1}{{1 + \frac{1}{x}}} = \frac{1}{{1 + x}} + \frac{x}{{1 + x}} = 1\),
因数列 \(\left\{ {{a_n}} \right\}\) 是正项等比数列,且 \({a_{10}} = 1\),则 \({a_1}{a_{19}} = {a_2}{a_{18}} = {a_3}{a_{17}} = \cdots = a_{10}^2 = 1\),
\(f({a_1}) + f({a_{19}}) = f({a_1}) + f(\frac{1}{{{a_1}}}) = 1\),同理\(f({a_2}) + f({a_{18}}) = f({a_3}) + f({a_{17}}) = \cdots = f({a_{10}}) + f({a_{10}}) = 1\),
令 \(S = f\left( {{a_1}} \right) + f\left( {{a_2}} \right) + f\left( {{a_3}} \right) + \cdots + f\left( {{a_{18}}} \right) + f\left( {{a_{19}}} \right)\),
又 \(S = f\left( {{a_{19}}} \right) + f\left( {{a_{18}}} \right) + f\left( {{a_{17}}} \right) + \cdots + f\left( {{a_2}} \right) + f\left( {{a_1}} \right)\),
则有 \(2S = 19\),\(S = \frac{{19}}{2}\),
所以 \(f\left( {{a_1}} \right) + f\left( {{a_2}} \right) + f\left( {{a_3}} \right) + \cdot \cdot \cdot + f\left( {{a_{18}}} \right) + f\left( {{a_{19}}} \right) = \frac{{19}}{2}\).
故答案为:\(\frac{{19}}{2}\)