Japan Mathematical Olympiad

The existence of a triangle with side lengths $a+f(b)$, $b+f(c)$, $c+f(a)$ is equivalent to the following conditions: $$\{\begin{array}{l}a+f(b)<b+f(c)+c+f(a),\\ b+f(c)<c+f(a)+a+f(b),\\ c+f(a)<a+f(b)+b+f(c).\end{array}$$ 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, $$(f(b)-b)-(f(a)-a)<f(c)+c.(\ast )$$ By considering the cases where $a=c=1$ and $b=c=1$ in $(\ast )$, 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 $(\ast )$, 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: $$\begin{array}{rlrlrlr}f(100)& \ge r-99,& f(101)& \ge r-100,& \dots ,& f(149)& \ge r-148,\\ f(150)& \ge k+150-r,& f(151)& \ge k+151-r,& \dots ,& f(199)& \ge k+199-r.\end{array}$$ Since all these inequalities hold, summing them up, we obtain $f(100)+f(101)+\cdots +f(199)\ge 50k+2550$.

On the other hand, define: $$g(x)=\{\begin{array}{ll}\frac{k}{2}+150-x& (100\le x\le 149)\\ \frac{k}{2}+x-149& (150\le x\le 199)\\ x+k& \text{(other cases).}\end{array}$$ 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 $50k+2550=102050$.