For how many positive integers $n$ less than or equal to $1000$ is \[(\sin t + i \cos t)^n=\sin nt + i \cos nt\] true for all real $t$?
To solve the problem, we need to determine for how many positive integers \( n \) less than or equal to \( 1000 \) the equation 
\[
(\sin t + i \cos t)^n = \sin(nt) + i \cos(nt)
\]
holds for all real \( t \).

1. **Initial Observation**:
   We start by testing \( t = 0 \). Substituting \( t = 0 \) into the equation, we get:
   \[
   (\sin 0 + i \cos 0)^n = \sin(0) + i \cos(0)
   \]
   Simplifying, we have:
   \[
   (0 + i \cdot 1)^n = 0 + i \cdot 1 \implies i^n = i
   \]
   This implies:
   \[
   i^n = i
   \]
   Since \( i^1 = i \), we need \( n \equiv 1 \pmod{4} \).

2. **Verification**:
   To verify, we factor out \( i \) from the right-hand side (RHS) of the original equation:
   \[
   \sin(nt) + i \cos(nt) = i(\cos(nt) - i \sin(nt)) = i \text{cis}(-nt)
   \]
   Similarly, for the left-hand side (LHS):
   \[
   (\sin t + i \cos t)^n = (i \cos t + \sin t)^n = (i \text{cis}(-t))^n = i^n \text{cis}(-nt)
   \]
   For the equality to hold, we need:
   \[
   i^n \text{cis}(-nt) = i \text{cis}(-nt)
   \]
   This implies:
   \[
   i^n = i
   \]
   Therefore, \( n \equiv 1 \pmod{4} \).

3. **Counting Solutions**:
   We need to count the number of positive integers \( n \) less than or equal to \( 1000 \) that satisfy \( n \equiv 1 \pmod{4} \). These numbers form an arithmetic sequence:
   \[
   1, 5, 9, \ldots, 997
   \]
   The general term of this sequence is given by:
   \[
   a_k = 1 + (k-1) \cdot 4 = 4k - 3
   \]
   Setting \( a_k \leq 1000 \):
   \[
   4k - 3 \leq 1000 \implies 4k \leq 1003 \implies k \leq \frac{1003}{4} \implies k \leq 250.75
   \]
   Since \( k \) must be an integer, the largest possible value for \( k \) is \( 250 \).

Thus, there are \( 250 \) positive integers \( n \) less than or equal to \( 1000 \) that satisfy the given condition.

The final answer is \(\boxed{250}\)