Let the coin be fair, so for each toss: $P(H)=\frac{1}{2}, \qquad P(T)=\frac{1}{2}.$

We need the probability of getting at least one head in $n$ tosses to be at least $0.9$.

Using the complement: $P(\text{at least one head}) = 1 - P(\text{no head}) = 1 - P(\text{all tails}).$

The probability of getting all tails in $n$ tosses is: $P(\text{all tails}) = \left(\frac{1}{2}\right)^n.$

Therefore, $P(\text{at least one head}) = 1 - \left(\frac{1}{2}\right)^n.$

Given: $1 - \left(\frac{1}{2}\right)^n \ge 0.9.$

Rearranging: $\left(\frac{1}{2}\right)^n \le 0.1.$

Now check integer values of $n$:

• For $n=3$: $\left(\frac{1}{2}\right)^3 = \frac{1}{8} = 0.125 > 0.1$ so this does not satisfy the condition.

• For $n=4$: $\left(\frac{1}{2}\right)^4 = \frac{1}{16} = 0.0625 \le 0.1$ so this does satisfy the condition.

Hence, the minimum value of $n$ is: $\boxed{4}.$