Among all polynomials $P(x)$ with integer coefficients for which $P(-10) = 145$ and $P(9) = 164$, compute the smallest possible value of $|P(0)|.$
1. **Given Conditions and Polynomial Properties:**
   We are given that \( P(x) \) is a polynomial with integer coefficients such that:
   \[
   P(-10) = 145 \quad \text{and} \quad P(9) = 164.
   \]
   We need to find the smallest possible value of \( |P(0)| \).

2. **Using Polynomial Properties:**
   It is well known that for any integers \( x \) and \( y \), and a polynomial \( P(x) \) with integer coefficients, the difference \( x - y \) divides \( P(x) - P(y) \). This is a consequence of the fact that \( P(x) - P(y) \) can be factored as \( (x-y)Q(x,y) \) where \( Q(x,y) \) is a polynomial with integer coefficients.

3. **Applying the Property:**
   - For \( x = 0 \) and \( y = -10 \):
     \[
     P(0) - P(-10) = P(0) - 145.
     \]
     Since \( 0 - (-10) = 10 \), it follows that:
     \[
     10 \mid (P(0) - 145).
     \]
     Therefore:
     \[
     P(0) \equiv 145 \pmod{10}.
     \]
     Simplifying, we get:
     \[
     P(0) \equiv 5 \pmod{10}.
     \]

   - For \( x = 0 \) and \( y = 9 \):
     \[
     P(0) - P(9) = P(0) - 164.
     \]
     Since \( 0 - 9 = -9 \), it follows that:
     \[
     9 \mid (P(0) - 164).
     \]
     Therefore:
     \[
     P(0) \equiv 164 \pmod{9}.
     \]
     Simplifying, we get:
     \[
     P(0) \equiv 2 \pmod{9}.
     \]

4. **Using the Chinese Remainder Theorem:**
   We now have the system of congruences:
   \[
   P(0) \equiv 5 \pmod{10}
   \]
   and
   \[
   P(0) \equiv 2 \pmod{9}.
   \]
   By the Chinese Remainder Theorem, there exists a unique solution modulo \( 90 \) (since \( 10 \) and \( 9 \) are coprime). We need to solve:
   \[
   P(0) \equiv 5 \pmod{10}
   \]
   and
   \[
   P(0) \equiv 2 \pmod{9}.
   \]

5. **Solving the System:**
   Let \( P(0) = 10k + 5 \) for some integer \( k \). Substituting into the second congruence:
   \[
   10k + 5 \equiv 2 \pmod{9}.
   \]
   Simplifying:
   \[
   10k + 5 \equiv 2 \pmod{9} \implies k + 5 \equiv 2 \pmod{9} \implies k \equiv -3 \pmod{9} \implies k \equiv 6 \pmod{9}.
   \]
   Therefore, \( k = 9m + 6 \) for some integer \( m \). Substituting back:
   \[
   P(0) = 10(9m + 6) + 5 = 90m + 65.
   \]
   Thus:
   \[
   P(0) \equiv 65 \pmod{90}.
   \]

6. **Minimizing \( |P(0)| \):**
   The general solution is \( P(0) = 90m + 65 \). To minimize \( |P(0)| \), we consider \( m = 0 \):
   \[
   P(0) = 65.
   \]
   We also consider \( m = -1 \):
   \[
   P(0) = 90(-1) + 65 = -25.
   \]
   The smallest absolute value is:
   \[
   |P(0)| = 25.
   \]

The final answer is \( \boxed{25} \).