Pam lists the four smallest positive prime numbers in increasing order. When she divides the positive integer $N$ by the first prime, the remainder is $1$. When she divides $N$ by the second prime, the remainder is $2$. When she divides $N$ by the third prime, the remainder is $3$. When she divides $N$ by the fourth prime, the remainder is $4$. Find the least possible value for $N$.
1. Identify the four smallest positive prime numbers: 2, 3, 5, and 7.
2. Set up the system of congruences based on the problem statement:
   \[
   \begin{cases}
   N \equiv 1 \pmod{2} \\
   N \equiv 2 \pmod{3} \\
   N \equiv 3 \pmod{5} \\
   N \equiv 4 \pmod{7}
   \end{cases}
   \]
3. Solve the first congruence:
   \[
   N \equiv 1 \pmod{2}
   \]
   This implies \(N\) is odd.

4. Solve the second congruence:
   \[
   N \equiv 2 \pmod{3}
   \]
   Since \(N\) is odd, we can write \(N = 3k + 2\) for some integer \(k\).

5. Solve the third congruence:
   \[
   N \equiv 3 \pmod{5}
   \]
   Substitute \(N = 3k + 2\) into the third congruence:
   \[
   3k + 2 \equiv 3 \pmod{5}
   \]
   Simplify:
   \[
   3k \equiv 1 \pmod{5}
   \]
   Find the multiplicative inverse of 3 modulo 5, which is 2 (since \(3 \cdot 2 \equiv 6 \equiv 1 \pmod{5}\)):
   \[
   k \equiv 2 \pmod{5}
   \]
   Therefore, \(k = 5m + 2\) for some integer \(m\).

6. Substitute \(k = 5m + 2\) back into \(N = 3k + 2\):
   \[
   N = 3(5m + 2) + 2 = 15m + 6 + 2 = 15m + 8
   \]
   Thus, \(N \equiv 8 \pmod{15}\).

7. Solve the fourth congruence:
   \[
   N \equiv 4 \pmod{7}
   \]
   Substitute \(N = 15m + 8\) into the fourth congruence:
   \[
   15m + 8 \equiv 4 \pmod{7}
   \]
   Simplify:
   \[
   15m \equiv -4 \pmod{7}
   \]
   Since \(15 \equiv 1 \pmod{7}\):
   \[
   m \equiv -4 \pmod{7}
   \]
   Simplify \(-4 \pmod{7}\):
   \[
   m \equiv 3 \pmod{7}
   \]
   Therefore, \(m = 7n + 3\) for some integer \(n\).

8. Substitute \(m = 7n + 3\) back into \(N = 15m + 8\):
   \[
   N = 15(7n + 3) + 8 = 105n + 45 + 8 = 105n + 53
   \]
   Thus, the least possible value for \(N\) is when \(n = 0\):
   \[
   N = 53
   \]

The final answer is \(\boxed{53}\).