Let $N>1$ be an integer. We are adding all remainders when we divide $N$ by all positive integers less than $N$. If this sum is less than $N$, find all possible values of $N$.
1. **Define the problem and notation:**
   Let \( N > 1 \) be an integer. We need to find all possible values of \( N \) such that the sum of the remainders when \( N \) is divided by all positive integers less than \( N \) is less than \( N \). Let \( r(N, q) \) be the remainder when \( N \) is divided by \( q \).

2. **Analyze the behavior of remainders:**
   For \( q \leq N/2 \), the remainders \( r(N, q) \) can vary significantly. However, for \( q > N/2 \), the remainders are more predictable. Specifically, for \( q > N/2 \), \( r(N, q) = N - q \).

3. **Sum of remainders for \( q > N/2 \):**
   Let \( M \) be the largest integer smaller than \( N/2 \). This means \( N = 2M + 1 \) or \( N = 2M + 2 \). We can write:
   \[
   \sum_{q=N-M}^{N-1} r(N, q) = \sum_{q=N-M}^{N-1} (N - q)
   \]
   This sum is equivalent to the sum of the first \( M \) positive integers:
   \[
   \sum_{k=1}^{M} k = \frac{M(M+1)}{2}
   \]

4. **Inequality constraints:**
   We know that the sum of all remainders must be less than \( N \):
   \[
   \sum_{q=1}^{N-1} r(N, q) < N
   \]
   Since the sum of remainders for \( q > N/2 \) is at least \( \frac{M(M+1)}{2} \), we have:
   \[
   \frac{M(M+1)}{2} < N
   \]

5. **Bounding \( M \):**
   Since \( N \approx 2M \) (either \( N = 2M + 1 \) or \( N = 2M + 2 \)), we can approximate \( N \) as \( 2M \). Thus, we need:
   \[
   \frac{M(M+1)}{2} < 2M
   \]
   Simplifying this inequality:
   \[
   M(M+1) < 4M
   \]
   \[
   M^2 + M < 4M
   \]
   \[
   M^2 - 3M < 0
   \]
   \[
   M(M - 3) < 0
   \]
   This inequality holds for \( 0 < M < 3 \). Since \( M \) is an integer, \( M \) can be 1 or 2.

6. **Possible values of \( N \):**
   - If \( M = 1 \), then \( N = 2M + 1 = 3 \) or \( N = 2M + 2 = 4 \).
   - If \( M = 2 \), then \( N = 2M + 1 = 5 \) or \( N = 2M + 2 = 6 \).

   Additionally, we need to check \( N = 2 \) since it is the smallest possible value greater than 1.

7. **Verification:**
   - For \( N = 2 \), the sum of remainders is \( 1 \), which is less than \( 2 \).
   - For \( N = 3 \), the sum of remainders is \( 1 + 2 = 3 \), which is not less than \( 3 \).
   - For \( N = 4 \), the sum of remainders is \( 1 + 2 + 3 = 6 \), which is not less than \( 4 \).
   - For \( N = 5 \), the sum of remainders is \( 1 + 2 + 3 + 4 = 10 \), which is not less than \( 5 \).
   - For \( N = 6 \), the sum of remainders is \( 1 + 2 + 3 + 4 + 5 = 15 \), which is not less than \( 6 \).

   Therefore, the only valid value is \( N = 2 \).

The final answer is \( \boxed{2} \).