1. Find all positive integers $n$, such that there exist $n$ consecutive positive integers whose sum is a perfect square.
For a positive integer $t$, let
$$
S(n, t)=\sum_{i=1}^{n}(t+i)=\frac{n(2 t+n+1)}{2} \text {. }
$$

When $n$ is odd, simply set $t=\frac{n-1}{2}$, and we have
$$
S(n, t)=n^{2} \text {. }
$$

When $n$ is even, let $n=2^{s} m\left(s \in \mathbf{Z}_{+}, m\right.$ be an odd number).
Notice that $2 t+n+1$ is odd.
Combining the definition of $S(n, t)$, we know that $2^{s-1} \| S(n, t)$.
Thus, when $s$ is even, $S(n, t)$ is not a perfect square;
When $s$ is odd, choose any odd number $x>n$, and set
$$
t=\frac{m x^{2}-n-1}{2} \in \mathbf{Z}_{+},
$$

then $S(n, t)=2^{s-1} m^{2} x^{2}$ is a perfect square.
In summary, the set of solutions for $n$ that satisfy the condition is $\left\{2^{s} m \mid m\right.$ is any odd number, $s$ is 0 or an odd number $\}$.