The existence of a triangle with side lengths $a + f(b)$, $b + f(c)$, $c + f(a)$ is equivalent to the following conditions:
$$\begin{cases} a + f(b) < b + f(c) + c + f(a), \\ b + f(c) < c + f(a) + a + f(b), \\ c + f(a) < a + f(b) + b + f(c). \end{cases}$$
By symmetry of $a, b, c$, the existence of a triangle with side lengths $a + f(b)$, $b + f(c)$, $c + f(a)$ for any positive integers $a, b, c$ is equivalent to the condition that for any positive integers $a, b, c$, we have $a + f(b) < b + f(c) + c + f(a)$, or equivalently,
$$(f(b) - b) - (f(a) - a) < f(c) + c. \quad (*)$$
By considering the cases where $a = c = 1$ and $b = c = 1$ in $(*)$, we can establish that $-2 < f(x) - x < 2f(1)$ holds for any positive integer $x$. Consequently, $f(x) - x$ has both a minimum and a maximum value. Denote the minimum and the maximum values as $m$ and $M$ respectively, and let $r = M - m$. Then, from $(*)$, we have $f(x) + x > r$, which implies $f(x) \ge r - x + 1$. Furthermore, let $k = f(34) - 34 = 1990$. Since $M \ge k$, it follows that $m = M - r \ge k - r$. Thus, for any positive integer $x$, we have $f(x) - x \ge k - r$, or equivalently, $f(x) \ge k + x - r$.
Therefore, we have:
$$\begin{aligned} f(100) &\ge r - 99, & f(101) &\ge r - 100, & \dots, & f(149) &\ge r - 148, \\ f(150) &\ge k + 150 - r, & f(151) &\ge k + 151 - r, & \dots, & f(199) &\ge k + 199 - r. \end{aligned}$$
Since all these inequalities hold, summing them up, we obtain $f(100) + f(101) + \dots + f(199) \ge 50k + 2550$.

On the other hand, define:
$$g(x) = \begin{cases} \frac{k}{2} + 150 - x & (100 \le x \le 149) \\ \frac{k}{2} + x - 149 & (150 \le x \le 199) \\ x + k & \text{(other cases).} \end{cases}$$
By considering each of three cases, it can be seen that for any positive integer $x$, $g(x) + x \ge \frac{k}{2} + 150$ and $\frac{k}{2} - 149 \le g(x) - x \le k$. Therefore, for any positive integers $a, b, c$,
$$(g(b) - b) - (g(a) - a) \le k - \left(\frac{k}{2} - 149\right) < \frac{k}{2} + 150 \le g(c) + c$$
holds, and since $g(34) = k + 34$, $g$ satisfies the conditions of the problem. For this function $g$, $g(100)+g(101)+\cdots+g(199) = 50k+2550$ holds, so the minimum to find is $50k+2550 = 102050$.