Since $\{x\} < 1$ holds for any $x$, we have $\{a_1\} + \{a_2\} + \cdots + \{a_{100}\} < 100$; also, by the relation between roots and coefficients, $-\sum_{i=1}^{100} a_i$ is the coefficient of the $x^{99}$ term of $f(x)$, and hence must be an integer, so
$$\{a_1\} + \{a_2\} + \cdots + \{a_{100}\} \le 99.$$
We now show that there exists an $f(x)$ such that $\{a_1\} + \{a_2\} + \cdots + \{a_{100}\} = 99$. Consider the function

$f(x) = x(x-2)(x-4)(x-6)\cdots(x-198) + (x-1)(x-3)\cdots(x-197)$

Substituting values for checking, it is easy to see that $f(-1)f(0), f(1)f(2), \cdots, f(197)f(198)$ are all less than 0, so by Newton's method for locating roots, $f(x)$ has a root in each of the 100 intervals $(-1, 0), (1, 2), (3, 4), \cdots, (197, 198)$. Therefore $f(x)$ has 100 real roots, satisfying the conditions required by the problem, and
$$[a_1] + [a_2] + \cdots + [a_{100}] = -1 + 1 + 3 + \cdots + 197$$
But the coefficient of the $x^{99}$ term of $f(x)$ is $-(2+4+6+\cdots+198)+1$, so by the relation between roots and coefficients, we have
$$a_1 + a_2 + \cdots + a_{100} = 2 + 4 + 6 + \cdots + 198 - 1$$
Therefore
$$$$
$$\begin{align*} \{a_1\} + \{a_2\} + \cdots + \{a_{100}\} &= (a_1 + a_2 + \cdots + a_{100}) - ([a_1] + [a_2] + \cdots + [a_{100}]) \\ &= (2 + 4 + 6 + \cdots + 198 - 1) \\ &\quad -(1 + 3 + 5 + \cdots + 197 - 1) \\ &= 99 \end{align*}$$