Find the number of positive integers less than or equal to $2019$ that are no more than $10$ away from a perfect square.
1. **Identify the range of perfect squares:**
   We need to find the perfect squares less than or equal to \(2019\). The largest integer \(n\) such that \(n^2 \leq 2019\) is \(n = 44\) because \(44^2 = 1936\) and \(45^2 = 2025\).

2. **Count the numbers close to each perfect square:**
   For each perfect square \(k^2\), the numbers within \(10\) units of \(k^2\) are \(k^2 - 10\) to \(k^2 + 10\). This gives us \(21\) numbers (including \(k^2\) itself).

3. **Handle the edge cases:**
   - For \(k = 1\) to \(k = 10\), the range \(k^2 - 10\) to \(k^2 + 10\) may include negative numbers or zero, which are not positive integers. We need to count only the positive integers in these ranges.
   - For \(k = 1\), the range is \(-9\) to \(11\), which includes \(1\) to \(11\) (11 numbers).
   - For \(k = 2\), the range is \(-6\) to \(14\), which includes \(1\) to \(14\) (14 numbers).
   - For \(k = 3\), the range is \(-1\) to \(19\), which includes \(1\) to \(19\) (19 numbers).
   - For \(k = 4\) to \(k = 10\), the range \(k^2 - 10\) to \(k^2 + 10\) includes all \(21\) numbers because \(k^2 - 10\) is positive.

4. **Count the numbers for \(k = 11\) to \(k = 44\):**
   Each of these perfect squares has exactly \(21\) numbers within \(10\) units of it.

5. **Sum the counts:**
   - For \(k = 1\) to \(k = 10\), we have:
     \[
     11 + 14 + 19 + 21 \times 7 = 11 + 14 + 19 + 147 = 191
     \]
   - For \(k = 11\) to \(k = 44\), we have:
     \[
     34 \times 21 = 714
     \]

6. **Add the counts together:**
   \[
   191 + 714 = 905
   \]

The final answer is \(\boxed{905}\).