Congratulations to the two student members who sent correct answers to this puzzle: Yudong Chen (University of Cambridge, UK) and Zhen Huang (Columbia University, USA). Here’s Anirban DasGupta’s solution:

Using the notation of the problem, the recorded values $Y_1, Y_2, \cdots$ are iid with
$E(Y_i) = \sum_{i = 1}^\infty \bigg (i\,c\,[F((i+1/2)\,c)-F((i-1/2)\,c)]\bigg )$,
where $c = \frac{1}{100}$, and $F(x) = 1-e^{-x/\lambda }$ is the CDF of the exponential distribution with mean $\lambda$. On using the formula
$\sum _{i = 1}^\infty i\,a^i = \frac{a}{(1-a)^2}$ for $|a| < 1$, we get $\bar{Y} \stackrel {P}{\rightarrow} E(Y_i) = \frac{e^{-\frac{1}{200\,\lambda }}}{100\,(1-e^{-\frac{1}{100\,\lambda }})} = \lambda - \frac{1}{24\,\times \,10^4\,\lambda }+\frac{7}{576\,\times \,10^9\,\lambda ^3} +O(\lambda ^{-5})$ as $\lambda \to \infty$. Thus, even for large $\lambda$, there is a small persistent bias in the estimator that we actually use in real life, and the real life estimator is inconsistent.