Let $b_i = \frac{a_i}{i}.$ We desire to show that: $$\prod_{i=1}^{99} (b_i + 1) \ge 100.$$ The condition says that $b_i + b_j \ge \frac1i + \frac1j$ for all $1 \le i < j \le 99.$ Suppose that $b_i < \frac1i$ for some $1 \le i \le 99.$ Let $c = \frac1i - b_i.$ Then, we've that $$(b_1 + 1)(b_2 + 1) \cdots (b_{99}+1) \ge \frac{1+\frac1i - c}{1+\frac1i + c} \cdot \prod_{j = 1}^{99} (1+\frac1j + c).$$ We clearly have that $\frac{1+\frac1i-c}{1+\frac1i + c} = \frac{1+\frac1i - c}{1+\frac1i} \cdot \frac{1+\frac1i}{1+\frac1i + c} \ge \frac{1}{(1+c)^2}.$ Now, observe that for $j \ge 90$, we have that $\frac{1+\frac1j+c}{1+\frac1j} \ge 1+ \frac{c}{2}$. Hence, as $(1+\frac{c}{2})^{10} > (1+c)^2,$ we're done. $\square$

Change 99 to 3, it is still true.