Let the initial number of inhabitants be $x^2$. Then we know $x^2+1000=y^2+1$, and thus $(y-x)(y+x)=999$ where $y$ is a positive integer. Additionally, $y^2+1+1000=z^2$, and thus $(z-y)(z+y)=1001$ where $z$ is a positive integer. Now we test cases, given $y-x$ and $y+x$ are positive integer factors that multiply to $999$, and we get that the four possibilities of $y$ are $500$, $168$, $60$, and $32$. Similarly testing cases for $z-y$ and $z+y$ multiplying to $1001$, we get that the four possibilities of $y$ here are $500$, $68$, $40$, and $32$. Clearly $y$ must equal either $500$ or $32$, but if $y=32$, $x=5$ from our first equation, and thus $x^2=25$ which clearly is less than $150$. So thus $y=500$, $x=499$, and $x^2=499^2=(500-1)^2=500^2-2 \cdot 500 +1 = 249001$ inhabitants.