Let $a$ and $b$ be positive integers such that the number $b^2 + (b +1)^2 +...+ (b + a)^2-3$ is multiple of $5$ and $a + b$ is odd. Calculate the digit of the units of the number $a + b$ written in decimal notation.
1. **Expression Simplification**:
   We start with the given expression:
   \[
   b^2 + (b+1)^2 + \cdots + (b+a)^2 - 3
   \]
   We need to determine when this expression is a multiple of 5.

2. **Sum of Squares**:
   The sum of squares from \(b^2\) to \((b+a)^2\) can be written as:
   \[
   \sum_{k=0}^{a} (b+k)^2
   \]
   Expanding this, we get:
   \[
   \sum_{k=0}^{a} (b^2 + 2bk + k^2) = (a+1)b^2 + 2b\sum_{k=0}^{a} k + \sum_{k=0}^{a} k^2
   \]
   Using the formulas for the sum of the first \(n\) integers and the sum of the first \(n\) squares, we have:
   \[
   \sum_{k=0}^{a} k = \frac{a(a+1)}{2} \quad \text{and} \quad \sum_{k=0}^{a} k^2 = \frac{a(a+1)(2a+1)}{6}
   \]
   Substituting these into our expression, we get:
   \[
   (a+1)b^2 + 2b \cdot \frac{a(a+1)}{2} + \frac{a(a+1)(2a+1)}{6}
   \]
   Simplifying, we obtain:
   \[
   (a+1)b^2 + b a(a+1) + \frac{a(a+1)(2a+1)}{6}
   \]

3. **Modulo 5 Analysis**:
   We need this expression minus 3 to be a multiple of 5:
   \[
   (a+1)b^2 + b a(a+1) + \frac{a(a+1)(2a+1)}{6} - 3 \equiv 0 \pmod{5}
   \]
   Since \(a\) and \(b\) are positive integers, we need to check the possible values modulo 5.

4. **Possible Values**:
   We are given that \(a + b\) is odd. We need to find \(a\) and \(b\) such that the above expression is a multiple of 5 and \(a + b\) is odd.

5. **Checking \(a \equiv 1 \pmod{5}\) and \(b \equiv 1 \pmod{5}\)**:
   If \(a \equiv 1 \pmod{5}\) and \(b \equiv 1 \pmod{5}\), then:
   \[
   a + b \equiv 1 + 1 \equiv 2 \pmod{5}
   \]
   Since \(a + b\) is odd, we can write:
   \[
   a + b \equiv 7 \pmod{10}
   \]

The final answer is \(\boxed{7}\).