Let $P(x)$ be a polynomial with integer coefficients and roots $1997$ and $2010$. Suppose further that $|P(2005)|<10$. Determine what integer values $P(2005)$ can get.
1. **Given Information and Polynomial Properties**:
   - Let \( P(x) \) be a polynomial with integer coefficients.
   - The roots of \( P(x) \) are \( 1997 \) and \( 2010 \).
   - We are given that \( |P(2005)| < 10 \).

2. **Using Polynomial Properties**:
   - For polynomials with integer coefficients, if \( a \) and \( b \) are integers, then \( a - b \) divides \( P(a) - P(b) \). This is a consequence of the fact that \( P(x) \) can be written as \( P(x) = (x - r_1)(x - r_2) \cdots (x - r_n) \) where \( r_i \) are the roots of the polynomial.

3. **Applying the Property**:
   - Since \( 1997 \) is a root, \( P(1997) = 0 \).
   - Since \( 2010 \) is a root, \( P(2010) = 0 \).
   - Therefore, \( 2005 - 1997 = 8 \) and \( 2005 - 2010 = -5 \).

4. **Divisibility Conditions**:
   - From \( 2005 - 1997 = 8 \), we have \( 8 \mid P(2005) - P(1997) \). Since \( P(1997) = 0 \), it follows that \( 8 \mid P(2005) \).
   - From \( 2005 - 2010 = -5 \), we have \( 5 \mid P(2005) - P(2010) \). Since \( P(2010) = 0 \), it follows that \( 5 \mid P(2005) \).

5. **Combining Divisibility Conditions**:
   - Since \( 8 \mid P(2005) \) and \( 5 \mid P(2005) \), the least common multiple of 8 and 5 is 40. Therefore, \( 40 \mid P(2005) \).

6. **Magnitude Condition**:
   - We are given \( |P(2005)| < 10 \). The only multiples of 40 that satisfy this condition are \( 0 \).

7. **Conclusion**:
   - Therefore, \( P(2005) = 0 \).

The final answer is \( \boxed{ P(2005) = 0 } \).