How many $5$-digit numbers $N$ (in base $10$) contain no digits greater than $3$ and satisfy the equality $\gcd(N,15)=\gcd(N,20)=1$? (The leading digit of $N$ cannot be zero.)

[i]Based on a proposal by Yannick Yao[/i]
1. **Understanding the constraints:**
   - $N$ is a 5-digit number in base 10.
   - $N$ contains no digits greater than 3, so each digit of $N$ can be 0, 1, 2, or 3.
   - The leading digit of $N$ cannot be zero.
   - $\gcd(N, 15) = 1$ and $\gcd(N, 20) = 1$.

2. **Analyzing the gcd conditions:**
   - $\gcd(N, 15) = 1$ implies $N$ is not divisible by 3 and 5.
   - $\gcd(N, 20) = 1$ implies $N$ is not divisible by 2 and 5.

3. **Implications for the digits of $N$:**
   - Since $N$ is not divisible by 2, the units digit must be odd (1 or 3).
   - Since $N$ is not divisible by 3, the sum of its digits must not be divisible by 3.
   - Since $N$ is not divisible by 5, the units digit cannot be 0 or 5. Given the digits allowed (0, 1, 2, 3), the units digit must be 1 or 3.

4. **Counting valid numbers:**
   - The units digit has 2 choices: 1 or 3.
   - The leading digit has 3 choices: 1, 2, or 3 (it cannot be 0).
   - The middle three digits can be any of 0, 1, 2, or 3, giving $4^3 = 64$ combinations.

5. **Ensuring the sum of digits is not divisible by 3:**
   - We need to count the combinations where the sum of the digits is not divisible by 3.
   - Let's denote the digits of $N$ as $d_1d_2d_3d_4d_5$, where $d_1$ is the leading digit and $d_5$ is the units digit.
   - The sum of the digits is $d_1 + d_2 + d_3 + d_4 + d_5$.

6. **Calculating the total number of valid numbers:**
   - For each choice of the units digit (1 or 3), we need to ensure the sum of the remaining digits is not divisible by 3.
   - We can use the principle of inclusion-exclusion to count the valid combinations.

7. **Final calculation:**
   - For each choice of the units digit (1 or 3), there are $3 \times 64 = 192$ combinations for the other digits.
   - We need to exclude the combinations where the sum of the digits is divisible by 3.
   - By symmetry and the properties of modular arithmetic, approximately one-third of the combinations will have a sum divisible by 3.
   - Therefore, we exclude $\frac{192}{3} = 64$ combinations for each choice of the units digit.

8. **Total valid numbers:**
   - For each units digit (1 or 3), there are $192 - 64 = 128$ valid combinations.
   - Since there are 2 choices for the units digit, the total number of valid numbers is $2 \times 128 = 256$.

The final answer is $\boxed{256}$.