By what least number must $1260$ be divided to make the quotient a perfect square?

Let's determine the prime factorization of $1260$:
$$1260=2\times 630=2^2\times 315=2^2\times 3\times 105=2^2\times 3^2\times 35=2^2\times 3^2\times 5^1\times 7^1$$

To make the quotient a perfect square, we must eliminate the prime factors that do not form pairs (i.e., those with odd exponents, which are $5$ and $7$).

$$\text{Required divisor}=5\times 7=35$$