In a row are written $1999$ numbers such that except the first and the last , each is equal to the sum of its neighbours. If the first number is $1$, find the last number. 

(V Senderov)
1. Let the sequence be denoted by \( a_1, a_2, a_3, \ldots, a_{1999} \).
2. Given that \( a_1 = 1 \) and for \( 2 \leq i \leq 1998 \), \( a_i = a_{i-1} + a_{i+1} \).
3. Let \( a_2 = x \). Then, we can express the subsequent terms using the given condition:
   - \( a_3 = a_2 - a_1 = x - 1 \)
   - \( a_4 = a_3 - a_2 = (x - 1) - x = -1 \)
   - \( a_5 = a_4 - a_3 = -1 - (x - 1) = -x \)
   - \( a_6 = a_5 - a_4 = -x - (-1) = -x + 1 \)
   - \( a_7 = a_6 - a_5 = (-x + 1) - (-x) = 1 \)
4. Notice that \( a_7 = a_1 = 1 \), indicating that the sequence repeats every 6 terms.
5. To find the 1999th term, we need to determine the position of 1999 in the repeating cycle:
   - \( 1999 \mod 6 = 5 \)
6. Therefore, the 1999th term corresponds to the 5th term in the repeating sequence, which is \( -x \).

The final answer is \( \boxed{ -x } \)